Chapter 11 Conic Sections (Additional Questions)

The correct answer is (D) Hyperbola.

When a plane intersects a double-napped cone, the shape formed by the intersection depends on the angle of the plane relative to the axis of the cone. These shapes are known as conic sections.

If the plane is parallel to the axis of the cone, it cuts through both nappes of the double cone. The resulting intersection consists of two separate, open curves that are mirror images of each other. This specific configuration defines a hyperbola.

Let's briefly consider the other options for comparison:

A Circle is formed when the plane is perpendicular to the axis of the cone and intersects only one nappe.

An Ellipse is formed when the plane intersects only one nappe and is not perpendicular to the axis, nor parallel to a generator of the cone.

A Parabola is formed when the plane intersects only one nappe and is parallel to a generator line of the cone.

Therefore, an intersection by a plane parallel to the axis of a double-napped cone results in a hyperbola.

The correct answer is (A) $(x-h)^2 + (y-k)^2 = r^2$.

The equation of a circle is derived from the distance formula. For any point $(x, y)$ on the circle, its distance from the center $(h, k)$ is equal to the radius $r$.

Using the distance formula, the distance between $(x, y)$ and $(h, k)$ is $\sqrt{(x-h)^2 + (y-k)^2}$.

Setting this distance equal to the radius $r$, we get:

Squaring both sides of the equation gives the standard form of the equation of a circle:

This equation represents all points $(x, y)$ that are exactly $r$ units away from the center $(h, k)$.

Option (B) is incorrect because the terms $(x+h)^2$ and $(y+k)^2$ would represent a circle centered at $(-h, -k)$.

Option (C) is a special case of the standard equation where the center is at the origin $(0, 0)$, i.e., $h=0$ and $k=0$.

Option (D) is incorrect as it is not the correct algebraic form for the distance squared.

The correct answer is (A) Center $(2, -3)$, Radius 4.

The general equation of a circle is given by $x^2 + y^2 + 2gx + 2fy + c = 0$.

The center of the circle is $(-g, -f)$ and the radius is $\sqrt{g^2 + f^2 - c}$.

Comparing the given equation $x^2 + y^2 - 4x + 6y - 3 = 0$ with the general equation, we have:

$2g = -4 \implies g = -2$

$2f = 6 \implies f = 3$

$c = -3$

Therefore, the center of the circle is:

Center $(h, k) = (-g, -f) = (-(-2), -(3)) = (2, -3)$.

The radius of the circle is:

Radius $r = \sqrt{g^2 + f^2 - c}$

$r = \sqrt{(-2)^2 + (3)^2 - (-3)}$

$r = \sqrt{4 + 9 + 3}$

$r = \sqrt{16}$

$r = 4$

Thus, the center of the circle is $(2, -3)$ and the radius is 4.

The correct answer is (B) $(x-a)^2 + (y-b)^2 = a^2 + b^2$.

We are given that the center of the circle is $(h, k) = (a, b)$.

The standard equation of a circle with center $(h, k)$ and radius $r$ is $(x-h)^2 + (y-k)^2 = r^2$.

Substituting the center $(a, b)$, the equation becomes:

$(x-a)^2 + (y-b)^2 = r^2$

We are also given that the circle passes through the origin $(0, 0)$. This means the point $(0, 0)$ lies on the circle.

The distance from the center $(a, b)$ to any point on the circle, including the origin $(0, 0)$, is the radius $r$.

Using the distance formula, the radius $r$ is the distance between $(a, b)$ and $(0, 0)$:

$r = \sqrt{(a-0)^2 + (b-0)^2}$

$r = \sqrt{a^2 + b^2}$

To get $r^2$, we square both sides:

$r^2 = (\sqrt{a^2 + b^2})^2$

$r^2 = a^2 + b^2$

Now, substitute this value of $r^2$ back into the standard equation of the circle with center $(a, b)$:

$(x-a)^2 + (y-b)^2 = r^2$

$(x-a)^2 + (y-b)^2 = a^2 + b^2$

This is the equation of the circle passing through the origin and having center at $(a, b)$.

Let's verify this by plugging in $(0, 0)$ into the derived equation:

$(0-a)^2 + (0-b)^2 = (-a)^2 + (-b)^2 = a^2 + b^2$.

The right side is $a^2 + b^2$. Since $a^2 + b^2 = a^2 + b^2$, the point $(0, 0)$ satisfies the equation, confirming that the circle passes through the origin.

The correct answer is (A) $y^2 = 4ax$.

We are given that the vertex of the parabola is at the origin $(0, 0)$.

The directrix is the line $x = -a$, where $a > 0$.

Since the directrix is a vertical line ($x = \text{constant}$), the axis of symmetry of the parabola is horizontal (parallel to the x-axis).

The directrix $x = -a$ is a vertical line located to the left of the y-axis because $a > 0$. A parabola opens away from its directrix. Therefore, this parabola opens to the right.

The standard equation of a parabola with vertex at the origin $(0, 0)$, a horizontal axis of symmetry, and opening to the right is $y^2 = 4px$, where $p$ is the distance from the vertex to the focus (and also the distance from the vertex to the directrix).

For this standard form, the directrix is $x = -p$.

We are given the directrix $x = -a$. Comparing this with $x = -p$, we find that $p = a$.

Substituting $p = a$ into the standard equation $y^2 = 4px$, we get the equation of the parabola:

$y^2 = 4ax$

Let's consider why the other options are incorrect:

(B) $y^2 = -4ax$: This parabola has vertex $(0,0)$ but opens to the left, as its directrix would be $x = a$.

(C) $x^2 = 4ay$: This parabola has vertex $(0,0)$, a vertical axis of symmetry, and opens upwards, with directrix $y = -a$.

(D) $x^2 = -4ay$: This parabola has vertex $(0,0)$, a vertical axis of symmetry, and opens downwards, with directrix $y = a$.

Therefore, the equation of the parabola with vertex $(0,0)$ and directrix $x = -a$ is indeed $y^2 = 4ax$.

The correct answer is (A) $(2, 0)$.

The given equation of the parabola is $y^2 = 8x$.

This equation is in the standard form of a parabola with vertex at the origin and a horizontal axis of symmetry, which is $y^2 = 4px$.

Comparing $y^2 = 8x$ with $y^2 = 4px$, we can equate the coefficients of $x$:

Solving for $p$:

Since the equation is of the form $y^2 = 4px$ with $p > 0$, the parabola opens to the right. The vertex is at the origin $(0, 0)$.

The focus of a parabola in the form $y^2 = 4px$ is located at $(p, 0)$.

Substituting the value of $p=2$, the focus is at $(2, 0)$.

The correct answer is (D) 12.

The given equation of the parabola is $x^2 = -12y$.

This equation is in the standard form of a parabola with vertex at the origin and a vertical axis of symmetry, which is $x^2 = -4ay$ (for a parabola opening downwards).

Comparing the given equation $x^2 = -12y$ with the standard form $x^2 = -4ay$, we can equate the coefficients of $y$:

Solving for $a$:

For a parabola in the form $x^2 = \pm 4ay$ or $y^2 = \pm 4ax$, the length of the latus rectum is given by $|4a|$.

In this case, the length of the latus rectum is $|-12|$ or $4a$.

Using $a=3$:

Length of latus rectum $= 4a = 4 \times 3 = 12$.

Thus, the length of the latus rectum of the parabola $x^2 = -12y$ is 12.

The correct answer is (A) Both A and R are true and R is the correct explanation of A.

Let's analyze the Assertion (A) and Reason (R).

Assertion (A): The eccentricity of an ellipse is less than 1.

This statement is a fundamental property of an ellipse. For an ellipse, the eccentricity $e$ always satisfies $0 \leq e < 1$. Circles are a special case of ellipses where $e=0$. So, Assertion (A) is True.

Reason (R): The eccentricity of a conic section is defined as the ratio of the distance of a point on the conic from the focus to its distance from the directrix.

This statement provides the definition of the eccentricity ($e$) of a conic section. For any point $P$ on a conic section, the distance from $P$ to the focus ($F$) divided by the distance from $P$ to the directrix ($L$) is a constant value, which is the eccentricity $e$. This can be written as:

This definition is correct for all conic sections (parabola, ellipse, hyperbola). So, Reason (R) is True.

Now, let's determine if Reason (R) is the correct explanation for Assertion (A).

The value of eccentricity $e$ determined by the ratio in Reason (R) is precisely what distinguishes different types of conic sections:

• If $e = 1$, the conic is a parabola.
• If $0 \leq e < 1$, the conic is an ellipse.
• If $e > 1$, the conic is a hyperbola.

The definition given in Reason (R) is the basis for why an ellipse has an eccentricity less than 1. The geometrical property of an ellipse is such that for any point on the curve, its distance from the focus is always less than its distance from the directrix. This directly leads to the ratio (eccentricity) being less than 1.

Therefore, Reason (R) correctly explains why the eccentricity of an ellipse (as defined by R) is less than 1.

Thus, both Assertion (A) and Reason (R) are true, and Reason (R) is the correct explanation of Assertion (A).

The correct answer is (B) (i)-(b), (ii)-(a), (iii)-(c).

Let's match each conic section with its standard equation, assuming the center is at the origin $(0,0)$ and the axes are along the coordinate axes:

• (i) Circle: The standard equation of a circle with center at the origin $(0,0)$ and radius $r$ is the set of all points $(x, y)$ such that the distance from $(x, y)$ to $(0,0)$ is $r$. Using the distance formula, this is $\sqrt{(x-0)^2 + (y-0)^2} = r$, which simplifies to $\sqrt{x^2 + y^2} = r$. Squaring both sides gives $x^2 + y^2 = r^2$. This matches equation (b).

• (ii) Ellipse: The standard equation of an ellipse centered at the origin with its axes along the coordinate axes is of the form $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$, where $a$ and $b$ are the lengths of the semi-major and semi-minor axes. This matches equation (a).

• (iii) Hyperbola: The standard equation of a hyperbola centered at the origin with its transverse axis along the x-axis is $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$. If the transverse axis is along the y-axis, the equation is $\frac{y^2}{b^2} - \frac{x^2}{a^2} = 1$. The given equation (c) $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$ represents a hyperbola. This matches equation (c).

Therefore, the correct matching is:

• (i) Circle - (b) $x^2 + y^2 = r^2$
• (ii) Ellipse - (a) $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$
• (iii) Hyperbola - (c) $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$

This corresponds to option (B).

The correct answer is (A) $\frac{x^2}{169} + \frac{y^2}{144} = 1$.

Given:

Foci of the ellipse are $(\pm 5, 0)$.

Vertices of the ellipse are $(\pm 13, 0)$.

To Find:

The equation of the ellipse.

Solution:

Since the foci are $(\pm 5, 0)$ and the vertices are $(\pm 13, 0)$, they lie on the x-axis. This indicates that the major axis of the ellipse is along the x-axis.

The center of the ellipse is the midpoint of the segment joining the foci or the vertices. The midpoint of $(\pm 5, 0)$ is $\left(\frac{5 + (-5)}{2}, \frac{0+0}{2}\right) = (0, 0)$. Similarly, the midpoint of $(\pm 13, 0)$ is $(0, 0)$.

So, the center of the ellipse is at the origin $(0, 0)$.

The standard equation of an ellipse with center at the origin and major axis along the x-axis is:

where $a$ is the length of the semi-major axis and $b$ is the length of the semi-minor axis.

For an ellipse with major axis along the x-axis, the vertices are located at $(\pm a, 0)$.

Given vertices are $(\pm 13, 0)$. Comparing this with $(\pm a, 0)$, we get:

The foci are located at $(\pm c, 0)$, where $c$ is the distance from the center to the foci.

Given foci are $(\pm 5, 0)$. Comparing this with $(\pm c, 0)$, we get:

For an ellipse, the relationship between $a$, $b$, and $c$ is given by $c^2 = a^2 - b^2$.

We need to find $b^2$. Rearranging the formula, we get $b^2 = a^2 - c^2$.

Substitute the values of $a$ and $c$:

$a^2 = 13^2 = 169$

$c^2 = 5^2 = 25$

$b^2 = 169 - 25$

$b^2 = 144$

Now, substitute the values of $a^2$ and $b^2$ into the standard equation of the ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$:

This is the equation of the ellipse.

The correct answer is (B) $4/5$.

The given equation of the ellipse is $\frac{x^2}{25} + \frac{y^2}{9} = 1$.

This equation is in the standard form $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ where the major axis is along the x-axis because the denominator under the $x^2$ term (25) is greater than the denominator under the $y^2$ term (9).

Comparing the given equation with the standard form, we have:

$a^2 = 25$

$b^2 = 9$

From $a^2 = 25$, we get $a = \sqrt{25} = 5$ (since $a$ represents half the length of the major axis, it must be positive).

From $b^2 = 9$, we get $b = \sqrt{9} = 3$ (since $b$ represents half the length of the minor axis, it must be positive).

The eccentricity $e$ of an ellipse is given by the formula $e = \frac{c}{a}$, where $c$ is the distance from the center to the focus. The relationship between $a$, $b$, and $c$ for an ellipse is $c^2 = a^2 - b^2$.

Calculate $c^2$:

$c^2 = a^2 - b^2$

$c^2 = 25 - 9$

$c^2 = 16$

Find $c$:

$c = \sqrt{16} = 4$ (since $c$ is a distance, we take the positive root).

Now, calculate the eccentricity $e$:

$e = \frac{c}{a}$

$e = \frac{4}{5}$

The eccentricity of the given ellipse is $4/5$. Since $4/5 < 1$, this confirms it is an ellipse.

The correct answer is (A) $\frac{x^2}{16} - \frac{y^2}{20} = 1$.

Given:

Foci of the hyperbola are $(\pm 6, 0)$.

Vertices of the hyperbola are $(\pm 4, 0)$.

To Find:

The equation of the hyperbola.

Solution:

Since the foci are $(\pm 6, 0)$ and the vertices are $(\pm 4, 0)$, they lie on the x-axis. This indicates that the transverse axis of the hyperbola is along the x-axis.

The center of the hyperbola is the midpoint of the segment joining the foci or the vertices. The midpoint of $(\pm 6, 0)$ is $\left(\frac{6 + (-6)}{2}, \frac{0+0}{2}\right) = (0, 0)$. Similarly, the midpoint of $(\pm 4, 0)$ is $(0, 0)$.

So, the center of the hyperbola is at the origin $(0, 0)$.

The standard equation of a hyperbola with center at the origin and transverse axis along the x-axis is:

where $a$ is the distance from the center to a vertex along the transverse axis, and $b$ is related to the length of the conjugate axis.

For a hyperbola with transverse axis along the x-axis, the vertices are located at $(\pm a, 0)$.

Given vertices are $(\pm 4, 0)$. Comparing this with $(\pm a, 0)$, we get:

The foci are located at $(\pm c, 0)$, where $c$ is the distance from the center to a focus.

Given foci are $(\pm 6, 0)$. Comparing this with $(\pm c, 0)$, we get:

For a hyperbola, the relationship between $a$, $b$, and $c$ is given by $c^2 = a^2 + b^2$.

We need to find $b^2$. Rearranging the formula, we get $b^2 = c^2 - a^2$.

Substitute the values of $a$ and $c$:

$a^2 = 4^2 = 16$

$c^2 = 6^2 = 36$

$b^2 = 36 - 16$

$b^2 = 20$

Now, substitute the values of $a^2$ and $b^2$ into the standard equation of the hyperbola $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$:

This is the equation of the hyperbola.

The correct answer is (C) $5/4$.

The given equation of the hyperbola is $\frac{x^2}{16} - \frac{y^2}{9} = 1$.

This equation is in the standard form of a hyperbola with the transverse axis along the x-axis, which is $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$.

Comparing the given equation with the standard form, we have:

$a^2 = 16$

$b^2 = 9$

From $a^2 = 16$, we get $a = \sqrt{16} = 4$ (since $a$ represents half the length of the transverse axis, it must be positive).

From $b^2 = 9$, we get $b = \sqrt{9} = 3$ (since $b$ represents half the length of the conjugate axis, it must be positive).

The eccentricity $e$ of a hyperbola is given by the formula $e = \frac{c}{a}$, where $c$ is the distance from the center to the focus. The relationship between $a$, $b$, and $c$ for a hyperbola is $c^2 = a^2 + b^2$.

Calculate $c^2$:

$c^2 = a^2 + b^2$

$c^2 = 16 + 9$

$c^2 = 25$

Find $c$:

$c = \sqrt{25} = 5$ (since $c$ is a distance, we take the positive root).

Now, calculate the eccentricity $e$:

$e = \frac{c}{a}$

$e = \frac{5}{4}$

The eccentricity of the given hyperbola is $5/4$. Note that for a hyperbola, the eccentricity $e > 1$, and $5/4 = 1.25$, which is indeed greater than 1.

The correct answer is (C) $y^2 = 8x$.

Given:

Shape of the satellite dish: Parabolic reflector.

Vertex of the parabola is at the origin $(0,0)$.

Axis of symmetry is along the positive x-axis.

Distance of the focus from the vertex is 2 meters.

To Find:

The equation of the parabola.

Solution:

We are given that the vertex of the parabola is at the origin $(0,0)$ and the axis of symmetry is along the positive x-axis.

The standard equation of a parabola with vertex at the origin and axis along the positive x-axis is:

$y^2 = 4px$

where $p$ is the distance from the vertex to the focus.

We are given that the focus is located 2 meters from the vertex. Therefore, the value of $p$ is:

Substitute the value of $p$ into the standard equation:

$y^2 = 4(2)x$

$y^2 = 8x$

This is the equation of the parabola.

Note: The information about the dish being 5 meters wide at the opening is not required to find the equation of the parabola, only the distance of the focus from the vertex is needed.

The correct answer is (A) 2 meters.

Given:

Vertex of the parabola is at the origin $(0,0)$.

Axis of symmetry is along the positive x-axis.

Distance of the focus from the vertex is 2 meters ($p=2$).

Equation of the parabola (from previous question) is $y^2 = 8x$.

To Find:

The distance from the vertex to the directrix.

Solution:

For a parabola with vertex at the origin $(0,0)$ and axis of symmetry along the x-axis, the standard form of the equation is $y^2 = 4px$, where $|p|$ is the distance from the vertex to the focus and also the distance from the vertex to the directrix.

From Question 14, we found that the equation of the parabola is $y^2 = 8x$.

Comparing $y^2 = 8x$ with the standard form $y^2 = 4px$, we have:

Solving for $p$:

The distance from the vertex to the directrix is $|p|$.

Distance from vertex to directrix $= |2| = 2$ meters.

Since the axis of symmetry is the positive x-axis and the vertex is at the origin, the directrix is a vertical line given by $x = -p$.

Directrix equation is $x = -2$. The distance from the vertex $(0,0)$ to the line $x = -2$ is indeed 2 units.

The correct answer is (D) $25/32$ meters.

Given:

The parabola has vertex at the origin $(0,0)$ and axis of symmetry along the positive x-axis.

The equation of the parabola is $y^2 = 8x$ (from Question 14).

The width of the dish is 5 meters, measured along a line perpendicular to the axis of symmetry.

To Find:

The distance from the vertex to the line where the width is measured.

Solution:

The width of the dish (5 meters) is measured perpendicular to the axis of symmetry (the x-axis). This means the two endpoints of the dish opening have y-coordinates that are equal in magnitude but opposite in sign.

Let the y-coordinates of the endpoints be $y_1$ and $y_2$. The total width is $|y_1 - y_2| = 5$. Since the axis is the x-axis and the vertex is at the origin, these points are $(x, y)$ and $(x, -y)$, where the width is $|y - (-y)| = |2y| = 5$.

So, $2|y| = 5$, which means $|y| = \frac{5}{2} = 2.5$. The y-coordinates of the endpoints are $y = 2.5$ and $y = -2.5$.

These endpoints lie on the parabola $y^2 = 8x$. We need to find the x-coordinate of these points. This x-coordinate represents the distance from the vertex (which is at $x=0$) to the line where the opening is.

Substitute $y = 2.5$ (or $y = -2.5$) into the equation of the parabola:

We can write $2.5$ as $\frac{5}{2}$:

Now, solve for $x$:

The x-coordinate of the points at the opening is $25/32$. Since the vertex is at $x=0$, the distance from the vertex to the line where the opening is located (the line $x = 25/32$) is $\frac{25}{32}$ meters.

The correct answer is (B) Square.

A conic section is a curve formed by the intersection of a plane with a double-napped right circular cone.

The main types of conic sections are:

• Circle (formed when the plane is perpendicular to the axis of the cone and does not pass through the vertex).

• Ellipse (formed when the plane intersects one nappe at an angle and is not perpendicular to the axis or parallel to a generator).

• Parabola (formed when the plane is parallel to a generator of the cone and intersects one nappe).

• Hyperbola (formed when the plane is parallel to the axis of the cone or intersects both nappes).

Looking at the options:

• Circle is a conic section.

• Parabola is a conic section.

• Hyperbola is a conic section.

• A Square is a four-sided polygon with equal sides and four right angles. It is a plane figure formed by straight lines and cannot be obtained by intersecting a cone with a plane.

Therefore, the Square is NOT a conic section.

The correct answer is (B) $\sqrt{2}$.

Solution:

A rectangular hyperbola (also known as an equilateral hyperbola) is a hyperbola in which the lengths of the semi-transverse axis ($a$) and the semi-conjugate axis ($b$) are equal.

So, for a rectangular hyperbola, we have:

The eccentricity $e$ of a hyperbola is defined as the ratio $e = \frac{c}{a}$ (or $e = \frac{c}{b}$ depending on the orientation, where $a$ is the semi-transverse axis). The distance $c$ from the center to the focus is related to $a$ and $b$ by the equation:

For a rectangular hyperbola, since $a = b$, we can substitute $b$ with $a$ in the equation for $c^2$:

Taking the square root of both sides (and considering positive values since $c$ and $a$ are lengths):

Now, calculate the eccentricity $e$ using the formula $e = \frac{c}{a}$:

Assuming $a \neq 0$:

The eccentricity of a rectangular hyperbola is always $\sqrt{2}$. Note that for any hyperbola, the eccentricity $e > 1$. $\sqrt{2} \approx 1.414$, which is indeed greater than 1.

The correct answer is (D) Both (B) and (C) are equivalent ways of representing it.

Solution:

The standard equation of an ellipse centered at the origin is $\frac{x^2}{\text{denominator}_1} + \frac{y^2}{\text{denominator}_2} = 1$.

The major axis lies along the axis corresponding to the term with the larger denominator.

We are given that the major axis is along the y-axis. This means the denominator under the $y^2$ term must be greater than the denominator under the $x^2$ term.

Let the semi-major axis length be denoted by $a$ and the semi-minor axis length be denoted by $b$. By definition, $a > b$.

When the major axis is along the y-axis, the standard form of the equation is:

In this form, $a^2$ (the square of the semi-major axis length) is under the $y^2$ term, and $b^2$ (the square of the semi-minor axis length) is under the $x^2$ term, where $a > b$.

Now let's examine the given options:

• (A) $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ (where $a > b$): This represents an ellipse with the major axis along the x-axis because the denominator under $x^2$ is larger ($a^2 > b^2$). So, this is incorrect for a major axis along the y-axis.

• (B) $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ (where $b > a$): In this option, the denominator under $y^2$ ($b^2$) is stated to be larger than the denominator under $x^2$ ($a^2$). This correctly indicates that the major axis is along the y-axis.

• (C) $\frac{x^2}{b^2} + \frac{y^2}{a^2} = 1$ (where $a$ is the semi-major axis length): In this option, $a^2$ (the square of the semi-major axis length) is explicitly placed under the $y^2$ term. Since the semi-major axis is defined as the larger semi-axis, $a$ is the larger value. Placing $a^2$ under $y^2$ means the major axis is along the y-axis. This is also a correct representation.

Comparing options (B) and (C):

In option (B), the variables $a$ and $b$ are used for the denominators, and the condition $b > a$ tells us the major axis is along the y-axis (because $b^2 > a^2$). The semi-major axis length would be $\sqrt{b^2} = b$, and the semi-minor axis length would be $\sqrt{a^2} = a$.

In option (C), the variables $a$ and $b$ are used such that $a$ is explicitly stated as the semi-major axis length and $b$ implicitly represents the semi-minor axis length (since $a > b$). By placing $a^2$ under the $y^2$ term and $b^2$ under the $x^2$ term, the equation $\frac{x^2}{b^2} + \frac{y^2}{a^2} = 1$ correctly represents an ellipse with the major axis along the y-axis (since $a^2 > b^2$).

Both options (B) and (C) describe the same ellipse type: one with the major axis along the y-axis. They just use the variables $a$ and $b$ differently to denote the relationship between the denominators and the axis lengths. If in option (B) we relabel $a$ as $b_{new}$ and $b$ as $a_{new}$ (where $a_{new} > b_{new}$), the equation becomes $\frac{x^2}{b_{new}^2} + \frac{y^2}{a_{new}^2} = 1$, which is the form in (C).

Therefore, both (B) and (C) are equivalent ways of representing the standard equation of an ellipse with the major axis along the y-axis and center at the origin.

The correct answer is (B) $x = -4$.

Given:

The equation of the parabola is $y^2 = 16x$.

To Find:

The equation of the directrix.

Solution:

The given equation of the parabola is $y^2 = 16x$.

This equation is in the standard form of a parabola with vertex at the origin $(0, 0)$ and axis of symmetry along the positive x-axis, which is $y^2 = 4px$.

Comparing the given equation $y^2 = 16x$ with the standard form $y^2 = 4px$, we equate the coefficients of $x$:

Solving for $p$:

For a parabola with vertex at the origin and axis of symmetry along the positive x-axis ($y^2 = 4px$, where $p > 0$), the directrix is a vertical line located at a distance $p$ from the vertex on the negative side of the x-axis.

The equation of the directrix is $x = -p$.

Substitute the value of $p=4$:

Thus, the equation of the directrix of the parabola $y^2 = 16x$ is $x = -4$.

The correct answer is (A) $x^2 + y^2 - 4y - 5 = 0$.

Given:

Endpoints of the diameter are $(x_1, y_1) = (2, -1)$ and $(x_2, y_2) = (-2, 5)$.

To Find:

The equation of the circle.

Solution:

We can find the equation of the circle using the endpoints of its diameter. The center of the circle is the midpoint of the diameter, and the radius is half the length of the diameter (or the distance from the center to either endpoint).

Step 1: Find the coordinates of the center of the circle.

The center $(h, k)$ is the midpoint of the segment connecting $(2, -1)$ and $(-2, 5)$.

$h = \frac{x_1 + x_2}{2} = \frac{2 + (-2)}{2} = \frac{0}{2} = 0$

$k = \frac{y_1 + y_2}{2} = \frac{-1 + 5}{2} = \frac{4}{2} = 2$

So, the center of the circle is $(0, 2)$.

Step 2: Find the radius of the circle.

The radius $r$ is the distance from the center $(0, 2)$ to one of the endpoints, say $(2, -1)$.

Using the distance formula:

$r = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$

$r = \sqrt{(2 - 0)^2 + (-1 - 2)^2}$

$r = \sqrt{(2)^2 + (-3)^2}$

$r = \sqrt{4 + 9}$

$r = \sqrt{13}$

The radius squared is $r^2 = (\sqrt{13})^2 = 13$.

Step 3: Write the equation of the circle.

The standard equation of a circle with center $(h, k)$ and radius $r$ is $(x-h)^2 + (y-k)^2 = r^2$.

Substitute the center $(0, 2)$ and $r^2 = 13$:

$(x - 0)^2 + (y - 2)^2 = 13$

$x^2 + (y^2 - 4y + 4) = 13$

$x^2 + y^2 - 4y + 4 - 13 = 0$

$x^2 + y^2 - 4y - 9 = 0$

Alternate Method: Using the diameter form of the circle's equation.

The equation of a circle with diameter endpoints $(x_1, y_1)$ and $(x_2, y_2)$ is given by $(x - x_1)(x - x_2) + (y - y_1)(y - y_2) = 0$.

Substitute the endpoints $(2, -1)$ and $(-2, 5)$:

$(x - 2)(x - (-2)) + (y - (-1))(y - 5) = 0$

$(x - 2)(x + 2) + (y + 1)(y - 5) = 0$

Expand the terms:

$(x^2 - (2)^2) + (y^2 - 5y + y - 5) = 0$

$x^2 - 4 + y^2 - 4y - 5 = 0$

$x^2 + y^2 - 4y - 9 = 0$

Both methods yield the equation $x^2 + y^2 - 4y - 9 = 0$.

Comparing this derived equation with the given options, it appears there is a discrepancy in the constant term. However, option (A) $x^2 + y^2 - 4y - 5 = 0$ is the only option that correctly represents the center $(0, 2)$ based on the coefficients of the $x$ and $y$ terms ($2g=0 \implies g=0$, $2f=-4 \implies f=-2$, Center $(-g, -f) = (0, 2)$).

Given the multiple-choice format, option (A) is the closest match to the correct equation, likely due to a typographical error in the constant term within the options. The calculated equation is $x^2 + y^2 - 4y - 9 = 0$.

The correct answer is (A) Both A and R are true and R is the correct explanation of A.

Let's analyze the Assertion (A) and Reason (R).

Assertion (A): A parabola has only one focus and one directrix.

This statement is a fundamental property of a parabola. Unlike ellipses and hyperbolas, which have two foci and two directrices (though they can be defined with one focus and one corresponding directrix), a parabola is characterized by a single focus and a single directrix. So, Assertion (A) is True.

Reason (R): A parabola is defined as the locus of a point that moves such that its distance from a fixed point (focus) is equal to its distance from a fixed line (directrix).

This statement provides the definition of a parabola. It is the set of all points $P$ in a plane that are equidistant from a fixed point $F$ (the focus) and a fixed line $L$ (the directrix) not passing through the focus. Mathematically, this definition is expressed as:

This ratio of distances is the eccentricity ($e$), and for a parabola, the eccentricity is always equal to 1 ($e=1$). This definition is correct. So, Reason (R) is True.

Now, let's determine if Reason (R) is the correct explanation for Assertion (A).

The definition in Reason (R) describes how a parabola is formed using one specific fixed point (the focus) and one specific fixed line (the directrix). The uniqueness of the focus and directrix used in this definition is precisely why the resulting curve, the parabola, has only one focus and one directrix. If there were multiple pairs of focus-directrix that defined the same parabola, the definition would reflect that. However, the standard definition uses a single pair.

Therefore, the definition provided in Reason (R) serves as the basis for the property stated in Assertion (A). Reason (R) correctly explains why a parabola is uniquely defined by and therefore possesses only one focus and one directrix.

Thus, both Assertion (A) and Reason (R) are true, and Reason (R) is the correct explanation of Assertion (A).

The correct answer is (A) ellipse.

Solution:

The definition provided in the question, "the locus of a point that moves such that the sum of its distances from two fixed points (foci) is constant," is the fundamental definition of an ellipse.

Let the two fixed points be foci $F_1$ and $F_2$, and let $P$ be any point on the locus. The definition states that $|PF_1| + |PF_2| = \text{constant}$. This constant is equal to $2a$, where $a$ is the length of the semi-major axis of the ellipse.

Let's consider the definitions of the other conic sections for comparison:

A parabola is the locus of a point such that its distance from a fixed point (focus) is equal to its distance from a fixed line (directrix).

A hyperbola is the locus of a point such that the absolute difference of its distances from two fixed points (foci) is constant.

A circle can be considered a special case of an ellipse where the two foci coincide at a single point (the center), and the sum of the distances from the two foci becomes twice the distance from the center, which is constant (the diameter). However, the primary definition of a circle is the locus of a point at a fixed distance (radius) from a single fixed point (center).

Based on the definition given, the shape formed is an ellipse.

The correct answer is (B) 10.

Given:

The equation of the ellipse is $\frac{x^2}{16} + \frac{y^2}{25} = 1$.

To Find:

The length of the major axis of the ellipse.

Solution:

The standard equation of an ellipse centered at the origin is $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$. The orientation of the major axis depends on whether $a^2$ or $b^2$ is the larger denominator. The larger denominator corresponds to the square of the semi-major axis length, and its position (under $x^2$ or $y^2$) indicates the direction of the major axis.

In the given equation, $\frac{x^2}{16} + \frac{y^2}{25} = 1$, we compare the denominators 16 and 25.

Since $25 > 16$, the denominator under the $y^2$ term is larger. This means the major axis is along the y-axis.

For an ellipse with the major axis along the y-axis, the standard form is $\frac{x^2}{b^2} + \frac{y^2}{a^2} = 1$, where $a$ is the length of the semi-major axis and $b$ is the length of the semi-minor axis ($a > b$).

Comparing $\frac{x^2}{16} + \frac{y^2}{25} = 1$ with $\frac{x^2}{b^2} + \frac{y^2}{a^2} = 1$:

We have $b^2 = 16$ and $a^2 = 25$.

From $a^2 = 25$, we find the length of the semi-major axis $a$:

The length of the major axis is $2a$.

Length of major axis $= 2 \times a = 2 \times 5 = 10$.

The length of the major axis is 10.

The correct answer is (C) $2ae$.

Given:

The equation of the hyperbola is $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$.

To Find:

The distance between the foci of the hyperbola.

Solution:

The given equation $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$ is the standard form of a hyperbola centered at the origin $(0,0)$ with the transverse axis along the x-axis.

For this type of hyperbola, the foci are located on the transverse axis (the x-axis) at coordinates $(\pm c, 0)$, where $c$ is the distance from the center to each focus.

The two foci are $F_1 = (-c, 0)$ and $F_2 = (c, 0)$.

The distance between the foci is the distance between these two points:

Distance $= \sqrt{(c - (-c))^2 + (0 - 0)^2}$

Distance $= \sqrt{(2c)^2 + 0^2}$

Distance $= \sqrt{4c^2}$

Distance $= |2c|$

Since $c$ is a distance, we take the positive value, so the distance between the foci is $2c$.

The eccentricity $e$ of a hyperbola is defined as the ratio $e = \frac{c}{a}$, where $a$ is the distance from the center to a vertex along the transverse axis.

From this definition, we can express $c$ in terms of $a$ and $e$:

Substitute this expression for $c$ into the distance between the foci:

Distance between foci $= 2c = 2(ae) = 2ae$.

Thus, the distance between the foci of the hyperbola $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$ is $2ae$.

The correct answer is (B) Parabola.

The eccentricity ($e$) is a parameter that measures how much a conic section deviates from being circular. It is defined for any point $P$ on the conic as the ratio of the distance from $P$ to a fixed point (the focus, $F$) to the distance from $P$ to a fixed line (the directrix, $L$).

The value of the eccentricity determines the type of conic section:

• If $e = 0$, the conic section is a Circle.

• If $0 \leq e < 1$, the conic section is an Ellipse (specifically, $0 < e < 1$ for a non-circular ellipse).

• If $e = 1$, the conic section is a Parabola.

• If $e > 1$, the conic section is a Hyperbola.

According to these definitions, the conic section with eccentricity equal to 1 is the Parabola.

The correct answer is (A) $x = r \cos \theta, y = r \sin \theta$.

Given:

The equation of the circle is $x^2 + y^2 = r^2$, centered at the origin with radius $r$.

To Find:

The standard parametric equations for this circle.

Solution:

Parametric equations represent the coordinates of points on a curve as functions of a single parameter. For a circle centered at the origin with radius $r$, we can use an angle $\theta$ (theta) as the parameter.

Consider a point $(x, y)$ on the circle in the xy-plane. If we draw a line segment from the origin to this point, let $\theta$ be the angle this segment makes with the positive x-axis. Using basic trigonometry in the right triangle formed by the point $(x, y)$, the origin $(0, 0)$, and the projection of the point onto the x-axis $(x, 0)$, we have:

$x = r \cos \theta$

$y = r \sin \theta$

These are the standard parametric equations for a circle $x^2 + y^2 = r^2$, where the parameter $\theta$ typically varies from $0$ to $2\pi$ to trace out the entire circle.

We can verify these equations by substituting them into the equation of the circle:

Using the fundamental trigonometric identity $\cos^2 \theta + \sin^2 \theta = 1$:

This matches the given equation of the circle.

The other options correspond to different curves or are not standard parametric forms:

• (B) $x = r \sec \theta, y = r \tan \theta$: These equations satisfy $x^2 - y^2 = r^2 (\sec^2 \theta - \tan^2 \theta) = r^2$, which is the equation of a hyperbola.

• (C) $x = r t, y = r t^2$: This parameterization, after eliminating $t$ (e.g., $t = x/r$, so $y = r(x/r)^2 = x^2/r$), represents a parabola ($y = \frac{1}{r}x^2$).

• (D) $x = r \theta, y = r \theta^2$: These equations do not represent a standard conic section in a simple form.

Thus, the standard parametric equations for the circle $x^2 + y^2 = r^2$ are $x = r \cos \theta$ and $y = r \sin \theta$.

The correct answer is (A) $x^2 + y^2 - 2x - 2y + 1 = 0$.

Given:

The circle passes through the points $A(1, 0)$ and $B(0, 1)$.

The center of the circle lies on the line $x+y=2$.

To Find:

The equation of the circle.

Solution:

Let the center of the circle be $C(h, k)$ and the radius be $r$. The equation of the circle is $(x-h)^2 + (y-k)^2 = r^2$.

Since the circle passes through points $A(1, 0)$ and $B(0, 1)$, the distance from the center to these points must be equal to the radius $r$.

Distance from $C(h, k)$ to $A(1, 0)$:

Distance from $C(h, k)$ to $B(0, 1)$:

Equating the two expressions for $r^2$:

Expand both sides:

Subtract $h^2 + k^2 + 1$ from both sides:

We are also given that the center $(h, k)$ lies on the line $x+y=2$. Substitute $(h, k)$ into the equation of the line:

Now we have a system of two equations with two variables:

Substitute $k=h$ into the second equation:

Since $h = k$, we have $k = 1$.

So, the center of the circle is $(h, k) = (1, 1)$.

Now, find the radius squared $r^2$ using either of the earlier equations for $r^2$. Let's use $(1-h)^2 + k^2 = r^2$:

The equation of the circle with center $(h, k) = (1, 1)$ and $r^2 = 1$ is:

Expand the equation:

Move the constant term to the left side:

This is the equation of the circle. Comparing this with the given options, it matches option (A).

The correct answer is (B) $2b^2/a$.

Given:

The equation of the ellipse is $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$, with the condition $a > b$.

To Find:

The length of the latus rectum of the ellipse.

Solution:

The equation of the ellipse is given as $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$.

The condition $a > b$ indicates that $a^2 > b^2$. Since $a^2$ is under the $x^2$ term, the major axis of the ellipse lies along the x-axis.

For an ellipse with the standard equation $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ where the major axis is along the x-axis (i.e., $a > b$), the length of the semi-major axis is $a$ and the length of the semi-minor axis is $b$.

The length of the latus rectum of an ellipse is given by the formula $\frac{2b^2}{a}$.

In this specific case, the values of $a$ and $b$ in the formula directly correspond to the $a$ and $b$ in the given equation, satisfying the condition $a > b$.

Therefore, the length of the latus rectum is $\frac{2b^2}{a}$.

The correct answer is (A) $2a$.

Given:

The equation of the hyperbola is $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$.

To Find:

The length of the transverse axis.

Solution:

The given equation $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$ is the standard form of a hyperbola centered at the origin $(0,0)$.

Since the $x^2$ term is positive and the $y^2$ term is negative, the transverse axis of the hyperbola lies along the x-axis.

For a hyperbola with its transverse axis along the x-axis, the vertices are located at $(\pm a, 0)$. The value of $a$ in the denominator of the positive term ($x^2/a^2$) represents the distance from the center to each vertex.

The two vertices are $V_1 = (-a, 0)$ and $V_2 = (a, 0)$.

The length of the transverse axis is the distance between these two vertices.

Length of transverse axis $= \text{Distance}(V_1, V_2)$

Length $= \sqrt{(a - (-a))^2 + (0 - 0)^2}$

Length $= \sqrt{(2a)^2 + 0^2}$

Length $= \sqrt{4a^2}$

Since $a$ in the standard equation is typically taken as a positive value representing a length, $\sqrt{a^2} = a$.

Length $= 2a$

Thus, the length of the transverse axis of the hyperbola $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$ is $2a$.

The correct answers are (A), (B), (C), and (D).

Solution:

The general equation of a second-degree curve in two variables is $Ax^2 + Bxy + Cy^2 + Dx + Ey + F = 0$.

For this equation to represent a circle, the following conditions must be met:

1. The coefficients of $x^2$ and $y^2$ must be equal and non-zero ($A = C \neq 0$).

2. The coefficient of the $xy$ term must be zero ($B = 0$).

If these conditions are met, the equation can be rewritten in the form $x^2 + y^2 + 2gx + 2fy + c = 0$ by dividing by $A$. The center of the circle is $(-g, -f)$ and the radius squared is $r^2 = g^2 + f^2 - c$.

A real circle exists if $r^2 > 0$. If $r^2 = 0$, the equation represents a single point (a degenerate circle). If $r^2 < 0$, the equation represents an imaginary circle. In the context of conic sections, "a circle" often includes the degenerate case unless specified otherwise.

Let's examine each option:

(A) $x^2 + y^2 = 9$

This is in the form $x^2 + y^2 = r^2$. Here, $A=1, C=1, B=0$. $A=C \neq 0$ and $B=0$.

The center is $(0, 0)$ and $r^2 = 9$. Since $r^2 = 9 > 0$, this represents a real circle with radius $r = 3$.

(B) $(x-1)^2 + (y+2)^2 = 4$

This is in the standard form $(x-h)^2 + (y-k)^2 = r^2$. Here, $h=1$, $k=-2$, and $r^2=4$.

Expanding this equation gives $x^2 - 2x + 1 + y^2 + 4y + 4 = 4$, which simplifies to $x^2 + y^2 - 2x + 4y + 1 = 0$. Here, $A=1, C=1, B=0$. $A=C \neq 0$ and $B=0$.

Since $r^2 = 4 > 0$, this represents a real circle with center $(1, -2)$ and radius $r = 2$.

(C) $x^2 + y^2 - 6x + 8y + 25 = 0$

This is in the general form. Here, $A=1, C=1, B=0$. $A=C \neq 0$ and $B=0$.

Comparing with $x^2 + y^2 + 2gx + 2fy + c = 0$, we have $2g = -6 \implies g = -3$, $2f = 8 \implies f = 4$, and $c = 25$.

The center is $(-g, -f) = (3, -4)$.

The radius squared is $r^2 = g^2 + f^2 - c = (-3)^2 + (4)^2 - 25 = 9 + 16 - 25 = 25 - 25 = 0$.

Since $r^2 = 0$, this equation represents a single point $(3, -4)$, which is a degenerate circle. It still fits the general form of a circle's equation.

(D) $2x^2 + 2y^2 - 4x + 6y - 10 = 0$

This is in the general form. Here, $A=2, C=2, B=0$. $A=C \neq 0$ and $B=0$.

Divide the entire equation by 2 to get the standard general form:

Comparing with $x^2 + y^2 + 2gx + 2fy + c = 0$, we have $2g = -2 \implies g = -1$, $2f = 3 \implies f = 3/2$, and $c = -5$.

The center is $(-g, -f) = (1, -3/2)$.

The radius squared is $r^2 = g^2 + f^2 - c = (-1)^2 + (3/2)^2 - (-5) = 1 + \frac{9}{4} + 5 = 6 + \frac{9}{4} = \frac{24+9}{4} = \frac{33}{4}$.

Since $r^2 = \frac{33}{4} > 0$, this represents a real circle with radius $r = \sqrt{33}/2$.

All four equations satisfy the conditions for representing a circle in the general form ($A=C \neq 0, B=0$). Options (A), (B), and (D) represent real circles with positive radii. Option (C) represents a point (a degenerate circle with zero radius). In a "Multiple Correct Answer(s)" question asking which equations represent "a circle", it is standard to include the degenerate case.

Therefore, all four options represent a circle.

The correct answer is (A) 0.

The eccentricity ($e$) of a conic section is a parameter that describes its shape. It is defined as the ratio of the distance from any point on the conic to a fixed point (the focus) to its distance from a fixed line (the directrix).

The value of eccentricity distinguishes the different types of conic sections:

• Circle: $e = 0$. A circle can be viewed as an ellipse where the two foci coincide at the center. In this case, the distance from any point on the circle to the focus (which is the center) is constant (the radius), while the directrix is considered to be at infinity.

• Ellipse: $0 \leq e < 1$. For a non-circular ellipse, $0 < e < 1$.

• Parabola: $e = 1$. The distance from any point on the parabola to the focus is equal to its distance from the directrix.

• Hyperbola: $e > 1$.

Therefore, the eccentricity of a circle is 0.

The correct answer is (B) Vertex $(-2, 1)$, Axis $y = 1$.

Given:

The equation of the parabola is $(y-1)^2 = 4(x+2)$.

To Find:

The vertex and axis of the parabola.

Solution:

The standard equation of a parabola with a horizontal axis of symmetry is $(y-k)^2 = 4p(x-h)$, where $(h, k)$ is the vertex and the axis of symmetry is the line $y = k$. The parabola opens to the right if $p > 0$ and to the left if $p < 0$.

The given equation is $(y-1)^2 = 4(x+2)$.

We can rewrite $x+2$ as $x - (-2)$.

Comparing $(y-1)^2 = 4(x - (-2))$ with the standard form $(y-k)^2 = 4p(x-h)$:

We can identify the values of $h$, $k$, and $4p$.

The vertex of the parabola is $(h, k)$.

Vertex $= (-2, 1)$.

The axis of symmetry for a parabola of the form $(y-k)^2 = 4p(x-h)$ is the horizontal line $y = k$.

Axis of symmetry $= y = 1$.

From $4p = 4$, we find $p = 1$. Since $p > 0$, the parabola opens to the right.

The focus is located at $(h+p, k) = (-2+1, 1) = (-1, 1)$.

The directrix is the vertical line $x = h-p = -2-1 = -3$.

Based on our calculations, the vertex is $(-2, 1)$ and the axis is $y = 1$. This matches option (B).

The correct answer is (A) $x x_1 + y y_1 = r^2$.

Given:

The equation of the circle is $x^2 + y^2 = r^2$.

The point of tangency is $(x_1, y_1)$, which lies on the circle.

To Find:

The equation of the tangent line to the circle at the point $(x_1, y_1)$.

Solution:

The equation of the tangent to a standard conic section at a point $(x_1, y_1)$ on the curve can be found by using a standard transformation rule on the equation of the conic section.

For a general second-degree equation $Ax^2 + Bxy + Cy^2 + Dx + Ey + F = 0$, the tangent at $(x_1, y_1)$ is obtained by replacing:

• $x^2$ with $xx_1$

• $y^2$ with $yy_1$

• $xy$ with $\frac{1}{2}(xy_1 + yx_1)$

• $x$ with $\frac{1}{2}(x+x_1)$

• $y$ with $\frac{1}{2}(y+y_1)$

• Constants ($F$) remain unchanged.

The given equation of the circle is $x^2 + y^2 = r^2$. We can write this as $x^2 + y^2 - r^2 = 0$.

Applying the transformation rules for the tangent at $(x_1, y_1)$:

• Replace $x^2$ with $xx_1$.

• Replace $y^2$ with $yy_1$.

• The constant term $-r^2$ remains $-r^2$.

Substitute these into the equation $x^2 + y^2 - r^2 = 0$:

Rearranging the equation, we get:

This is the equation of the tangent line to the circle $x^2 + y^2 = r^2$ at the point $(x_1, y_1)$.

The correct answer is (B) $(\pm 8, 0)$.

Given:

The equation of the ellipse is $\frac{x^2}{100} + \frac{y^2}{36} = 1$.

To Find:

The foci of the ellipse.

Solution:

The given equation of the ellipse is $\frac{x^2}{100} + \frac{y^2}{36} = 1$.

This equation is in the standard form $\frac{x^2}{A^2} + \frac{y^2}{B^2} = 1$.

We compare the denominators: $100$ and $36$. Since $100 > 36$, the larger denominator is under the $x^2$ term. This means the major axis of the ellipse lies along the x-axis, and the center is at the origin $(0,0)$.

For an ellipse with the major axis along the x-axis, the standard form is $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$, where $a$ is the length of the semi-major axis and $b$ is the length of the semi-minor axis, and $a > b$.

Comparing $\frac{x^2}{100} + \frac{y^2}{36} = 1$ with $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$:

We have $a^2 = 100$ and $b^2 = 36$.

From $a^2 = 100$, the length of the semi-major axis is $a = \sqrt{100} = 10$.

From $b^2 = 36$, the length of the semi-minor axis is $b = \sqrt{36} = 6$.

Note that $a = 10$ and $b = 6$, so $a > b$, which is consistent with the major axis being along the x-axis.

The foci of an ellipse with the major axis along the x-axis and center at the origin are located at $(\pm c, 0)$, where $c$ is related to $a$ and $b$ by the equation $c^2 = a^2 - b^2$.

Calculate $c^2$:

Find $c$:

The foci are located at $(\pm c, 0)$.

Foci $= (\pm 8, 0)$.

The correct answer is (B) $x = \pm a/e$.

Given:

The equation of the hyperbola is $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$.

To Find:

The equation of the directrices.

Solution:

The given equation $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$ is the standard form of a hyperbola centered at the origin $(0,0)$ with the transverse axis along the x-axis.

For a hyperbola with the transverse axis along the x-axis, the directrices are vertical lines perpendicular to the transverse axis.

The equation of the directrices for a hyperbola in the form $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$ is given by $x = \pm \frac{a}{e}$, where $a$ is the distance from the center to a vertex and $e$ is the eccentricity of the hyperbola.

The eccentricity $e$ is related to $a$ and $b$ by the equation $e = \frac{c}{a}$, where $c^2 = a^2 + b^2$.

The foci are located at $(\pm c, 0)$. The distance from the center to the foci is $c = ae$. The equations of the directrices are given by $x = \pm \frac{a^2}{c}$.

Substituting $c = ae$:

Thus, the equations of the directrices are $x = \frac{a}{e}$ and $x = -\frac{a}{e}$.

Given the parametric equations:

From equation (i), we can write:

From equation (ii), we can write:

We use the fundamental trigonometric identity:

Substitute the expressions for $\cos \theta$ and $\sin \theta$ from equations (iii) and (iv) into equation (v):

$\left(\frac{x}{a}\right)^2 + \left(\frac{y}{b}\right)^2 = 1$

$\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$

This equation is the standard form of the equation of an ellipse centered at the origin.

If $a=b$, the equation reduces to $\frac{x^2}{a^2} + \frac{y^2}{a^2} = 1$, which is $x^2 + y^2 = a^2$, representing a circle (a special case of an ellipse).

However, the general parametric form $x = a \cos \theta, y = b \sin \theta$ represents an ellipse with semi-axes $a$ and $b$ along the x and y directions, respectively.

Therefore, the parametric equations represent an ellipse.

The correct option is (C) An ellipse.

Let's analyze the Assertion (A) and Reason (R).

Assertion (A): The standard equation of a parabola with vertex at the origin $(0,0)$ and axis along the positive y-axis is $x^2 = 4ay$ (where $a > 0$).

A parabola whose axis is along the positive y-axis and vertex is at the origin indeed has the standard equation $x^2 = 4ay$. The condition $a > 0$ ensures that the parabola opens upwards along the positive y-axis.

Therefore, Assertion (A) is true.

Reason (R): In this case, the focus is $(0, a)$ and the directrix is $y = -a$.

For a parabola with the standard equation $x^2 = 4ay$ (where $a > 0$):

The vertex is at the origin $(0,0)$.

The axis of symmetry is the y-axis.

Since $a > 0$, the parabola opens upwards.

The focus is located on the axis of symmetry at a distance of $a$ units from the vertex in the direction the parabola opens. Thus, the focus is at $(0, a)$.

The directrix is a line perpendicular to the axis of symmetry, located at a distance of $a$ units from the vertex on the opposite side of the focus. Since the axis is the y-axis and the vertex is $(0,0)$, the directrix is the horizontal line $y = -a$.

Therefore, Reason (R) is true.

Now, let's consider if Reason (R) is the correct explanation for Assertion (A).

The definition of a parabola is the locus of points such that the distance from a fixed point (the focus) is equal to the distance from a fixed line (the directrix).

Using the focus $(0, a)$ and the directrix $y = -a$, let $(x, y)$ be any point on the parabola.

Distance from $(x, y)$ to the focus $(0, a)$ is $\sqrt{(x-0)^2 + (y-a)^2} = \sqrt{x^2 + (y-a)^2}$.

Distance from $(x, y)$ to the directrix $y = -a$ (or $y+a=0$) is $\frac{|y - (-a)|}{\sqrt{0^2 + 1^2}} = |y+a|$.

Equating the distances:

$\sqrt{x^2 + (y-a)^2} = |y+a|$

Squaring both sides:

$x^2 + (y-a)^2 = (y+a)^2$

$x^2 + y^2 - 2ay + a^2 = y^2 + 2ay + a^2$

Subtract $y^2 + a^2$ from both sides:

$x^2 - 2ay = 2ay$

Add $2ay$ to both sides:

$x^2 = 4ay$

This derivation shows that the standard equation $x^2 = 4ay$ is derived directly from the given focus $(0, a)$ and directrix $y = -a$. Therefore, Reason (R) correctly explains Assertion (A).

Both Assertion (A) and Reason (R) are true, and Reason (R) is the correct explanation of Assertion (A).

The correct option is (A) Both A and R are true and R is the correct explanation of A.

Conic sections are the curves obtained by intersecting a plane with a double-napped right circular cone.

The type of conic section formed depends on the angle between the plane and the axis of the cone, relative to the angle of the cone's generator.

Let $\beta$ be the angle between the intersecting plane and the axis of the cone, and let $\alpha$ be the semi-vertical angle of the cone (the angle between the generator and the axis).

If the plane is perpendicular to the axis of the cone, the angle between the plane and the axis is $90^\circ$. So, $\beta = 90^\circ$.

In this case, since $\beta = 90^\circ > \alpha$ and the plane does not pass through the vertex:

The intersection is a Circle.

For reference, other cases are:

- If $\alpha < \beta < 90^\circ$, the intersection is an Ellipse.

- If $\beta = \alpha$, the intersection is a Parabola.

- If $0 \leq \beta < \alpha$, the intersection is a Hyperbola.

- If the plane passes through the vertex, the intersections can be a point, a line, or two intersecting lines (degenerate cases).

According to the given condition, the plane is perpendicular to the axis ($\beta = 90^\circ$) and does not pass through the vertex.

This results in a Circle.

The correct option is (A) Circle.

Let's analyze each given equation to identify the type of conic section it represents.

(A) $x^2 - y^2 = 0$

This equation can be factored as $(x - y)(x + y) = 0$.

This means either $x - y = 0$ or $x + y = 0$.

These are the equations of two straight lines $y = x$ and $y = -x$ passing through the origin. This is a degenerate hyperbola.

(B) $x^2 + y^2 = 1$

This equation is in the standard form $x^2 + y^2 = r^2$.

It represents a circle centered at the origin $(0,0)$ with radius $r=1$.

(C) $y^2 = 4x$

This equation is in the standard form $y^2 = 4ax$.

It represents a parabola with its vertex at the origin $(0,0)$ and axis along the positive x-axis.

(D) $\frac{x^2}{4} - \frac{y^2}{9} = 1$

This equation is in the standard form $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$.

Here, $a^2 = 4$ and $b^2 = 9$. This is the equation of a hyperbola centered at the origin $(0,0)$ with its transverse axis along the x-axis.

Comparing the given options with the standard forms of conic sections, the equation $\frac{x^2}{4} - \frac{y^2}{9} = 1$ represents a non-degenerate hyperbola.

The correct option is (D) $\frac{x^2}{4} - \frac{y^2}{9} = 1$.

The concept of a latus rectum is typically defined for parabolas, ellipses, and hyperbolas. For these conic sections, the latus rectum is a chord passing through a focus and perpendicular to the axis (or major axis).

A circle can be considered a special case of an ellipse where the eccentricity $e = 0$ and the two foci coincide at the center of the circle.

The equation of a circle centered at the origin with radius $r$ is $x^2 + y^2 = r^2$.

The center (which acts as the single focus for a circle) is at $(0,0)$.

If we extend the definition of a latus rectum to a circle, it would be a chord passing through the focus (center) and perpendicular to an axis (any diameter). A chord passing through the center of a circle is a diameter.

Consider a diameter along the x-axis. The line perpendicular to this diameter passing through the center $(0,0)$ is the y-axis. The points of intersection of the y-axis ($x=0$) with the circle $x^2 + y^2 = r^2$ are $0^2 + y^2 = r^2$, so $y^2 = r^2$, which gives $y = \pm r$. The points are $(0, r)$ and $(0, -r)$. The length of the chord connecting these two points is $r - (-r) = 2r$.

Similarly, considering a diameter along the y-axis, the perpendicular line through the center is the x-axis ($y=0$). The intersection points are $(\pm r, 0)$, and the chord length is $r - (-r) = 2r$.

Thus, if we interpret the latus rectum of a circle as a chord passing through the center (focus) perpendicular to any diameter (axis), its length is equal to the diameter of the circle.

The length of the diameter of a circle with radius $r$ is $2r$.

While some stricter definitions of latus rectum for conics with $e > 0$ might lead to the conclusion that it is undefined for a circle ($e=0$), given the options provided, $2r$ is the most plausible intended answer based on extending the geometric interpretation.

The correct option is (B) $2r$.

The given equation of the circle is $x^2 + y^2 = r^2$.

This represents a circle centered at the origin $(0,0)$ with a radius of $r$.

Let $(x_1, y_1)$ be a point in the plane.

The distance of the point $(x_1, y_1)$ from the center of the circle $(0,0)$ is given by the distance formula:

$d = \sqrt{(x_1 - 0)^2 + (y_1 - 0)^2} = \sqrt{x_1^2 + y_1^2}$

For the point $(x_1, y_1)$ to lie inside the circle, its distance from the center must be less than the radius $r$.

So, we must have $d < r$.

Substituting the expression for $d$:

$\sqrt{x_1^2 + y_1^2} < r$

Since both sides of the inequality are non-negative (distance and radius), we can square both sides without changing the direction of the inequality:

$(\sqrt{x_1^2 + y_1^2})^2 < r^2$

$x_1^2 + y_1^2 < r^2$

The condition for the point $(x_1, y_1)$ to lie inside the circle $x^2 + y^2 = r^2$ is $x_1^2 + y_1^2 < r^2$.

Let's look at the other options for completeness:

If $x_1^2 + y_1^2 = r^2$, the point $(x_1, y_1)$ lies on the circle.

If $x_1^2 + y_1^2 > r^2$, the point $(x_1, y_1)$ lies outside the circle.

The correct option is (B) $x_1^2 + y_1^2 < r^2$.

Given:

Focus of the parabola, $F = (0, 6)$

Directrix of the parabola, $L$: $y = -6$

To Find:

The equation of the parabola.

Solution:

A parabola is defined as the locus of all points $(x, y)$ that are equidistant from a fixed point (the focus) and a fixed line (the directrix).

Let $P(x, y)$ be any point on the parabola.

The distance from $P$ to the focus $F(0, 6)$ is $PF$.

$PF = \sqrt{(x - 0)^2 + (y - 6)^2} = \sqrt{x^2 + (y - 6)^2}$

The distance from $P$ to the directrix $L$ ($y = -6$ or $y + 6 = 0$) is $PD$.

The formula for the distance from a point $(x_1, y_1)$ to a line $Ax + By + C = 0$ is $\frac{|Ax_1 + By_1 + C|}{\sqrt{A^2 + B^2}}$.

For the point $(x, y)$ and the line $0x + 1y + 6 = 0$, the distance is:

$PD = \frac{|0 \cdot x + 1 \cdot y + 6|}{\sqrt{0^2 + 1^2}} = \frac{|y + 6|}{\sqrt{1}} = |y + 6|$

By the definition of a parabola, $PF = PD$.

So, $\sqrt{x^2 + (y - 6)^2} = |y + 6|$

Squaring both sides to eliminate the square root and absolute value:

$(\sqrt{x^2 + (y - 6)^2})^2 = (|y + 6|)^2$

$x^2 + (y - 6)^2 = (y + 6)^2$

Expand the squared terms:

$x^2 + (y^2 - 12y + 36) = (y^2 + 12y + 36)$

Subtract $y^2 + 36$ from both sides of the equation:

$x^2 - 12y = 12y$

Add $12y$ to both sides:

$x^2 = 12y + 12y$

$x^2 = 24y$

This is the equation of the parabola.

Alternatively, the standard equation of a parabola with vertex at the origin $(0,0)$ and axis along the positive y-axis is $x^2 = 4ay$, where the focus is $(0, a)$ and the directrix is $y = -a$.

Given focus is $(0, 6)$ and directrix is $y = -6$. Comparing with the standard form, we find that $a = 6$.

Substituting $a = 6$ into the standard equation $x^2 = 4ay$, we get $x^2 = 4(6)y$, which simplifies to $x^2 = 24y$.

The equation of the parabola is $x^2 = 24y$.

Comparing this with the given options, it matches option (C).

The correct option is (C) $x^2 = 24y$.

The equation of the parabola is given by $y^2 = 4ax$.

We need to find the parametric equations for $x$ and $y$ in terms of a parameter (usually $t$ or $\theta$) such that substituting them into the parabola's equation satisfies the equation.

Let's test the given options by substituting the parametric forms for $x$ and $y$ into the equation $y^2 = 4ax$.

(A) $x = at^2, y = 2at$

Substitute these into $y^2 = 4ax$:

Left side of the equation: $y^2 = (2at)^2 = 4a^2t^2$

Right side of the equation: $4ax = 4a(at^2) = 4a^2t^2$

Since the Left Side equals the Right Side ($4a^2t^2 = 4a^2t^2$), these parametric equations satisfy the equation of the parabola $y^2 = 4ax$.

(B) $x = 2at, y = at^2$

Substitute these into $y^2 = 4ax$:

Left side of the equation: $y^2 = (at^2)^2 = a^2t^4$

Right side of the equation: $4ax = 4a(2at) = 8a^2t$

In general, $a^2t^4 \neq 8a^2t$. For example, if $t=1$, $a^2 \neq 8a^2$ unless $a=0$, which would not be a parabola. So, these are not the correct parametric equations for $y^2 = 4ax$. (Note: these are the parametric equations for $x^2=4ay$).

(C) $x = a \cos \theta, y = b \sin \theta$

These are the standard parametric equations for an ellipse centered at the origin, given by the Cartesian equation $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$.

(D) $x = a \sec \theta, y = b \tan \theta$

These are the standard parametric equations for a hyperbola centered at the origin with the transverse axis along the x-axis, given by the Cartesian equation $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$.

Based on our verification, the parametric equations $x = at^2, y = 2at$ correctly represent the parabola $y^2 = 4ax$.

The correct option is (A) $x = at^2, y = 2at$.

The equation of the ellipse is given by $\frac{x^2}{25} + \frac{y^2}{16} = 1$.

This equation is in the standard form $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ or $\frac{x^2}{b^2} + \frac{y^2}{a^2} = 1$.

We compare the given equation with the standard form.

We have $\frac{x^2}{25} + \frac{y^2}{16} = 1$.

Here, the denominator under the $x^2$ term is $25$, and the denominator under the $y^2$ term is $16$.

Since $25 > 16$, the major axis of the ellipse is along the x-axis.

For an ellipse with the major axis along the x-axis, the standard form is $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$, where $a > b$.

Comparing $\frac{x^2}{25} + \frac{y^2}{16} = 1$ with this form, we get:

From equation (i):

$a = \sqrt{25} = 5$ (since $a$ is a length, it is positive)

$a$ is the length of the semi-major axis.

From equation (ii):

$b = \sqrt{16} = 4$ (since $b$ is a length, it is positive)

$b$ is the length of the semi-minor axis.

The length of the major axis is $2a$.

The length of the minor axis is $2b$.

We are asked to find the length of the minor axis.

Length of minor axis $= 2b = 2 \times 4 = 8$.

The length of the minor axis is 8.

Comparing this with the given options, it matches option (C).

The correct option is (C) 8.

Definition of a Hyperbola:

A hyperbola is defined as the locus of all points in a plane such that the absolute difference of their distances from two fixed points (called the foci) in the plane is a constant positive value.

Let the two foci be $F_1$ and $F_2$, and let $P$ be any point on the hyperbola.

According to the definition, $|PF_1 - PF_2| = \text{constant}$.

For a standard hyperbola, this constant difference is equal to $2a$, where $a$ is the length of the semi-transverse axis.

The transverse axis is the line segment connecting the two vertices of the hyperbola. Its length is $2a$.

Thus, the difference of the distances of any point on the hyperbola from the two foci is constant and equal to the length of the transverse axis.

The correct option is (A) Transverse axis.

Let's analyze the Assertion (A) and Reason (R).

Assertion (A): A circle is a special case of an ellipse where the two foci coincide.

The definition of an ellipse is the locus of points where the sum of the distances from two fixed points (foci, $F_1$ and $F_2$) is constant, say $2a$. That is, $|PF_1| + |PF_2| = 2a$ for any point $P$ on the ellipse.

If the two foci $F_1$ and $F_2$ coincide at a single point, say $C$, then the sum of the distances becomes $|PC| + |PC| = 2|PC|$.

So, $2|PC| = 2a$, which simplifies to $|PC| = a$.

This equation states that the distance of any point $P$ from the fixed point $C$ is constant and equal to $a$. This is precisely the definition of a circle with center $C$ and radius $a$.

Therefore, a circle is indeed a special case of an ellipse where the two foci coincide. Assertion (A) is true.

Reason (R): In a circle, the distance of any point from the center (which is the coinciding foci) is constant (the radius), which fits the definition of an ellipse.

As explained above, when the foci of an ellipse coincide, they form the center of a circle. The definition of a circle is that all points on the curve are equidistant from the center, and this distance is the radius. The statement in Reason (R) accurately describes this property of a circle.

Furthermore, this property ($|PC| = r$) is derived from the ellipse definition when the foci coincide ($|PF_1| + |PF_2| = 2a$ becomes $2|PC| = 2a$, so $|PC| = a$, where $a$ is the radius $r$). The reason explains the geometric consequence of the coinciding foci and how it leads to the definition of a circle.

Therefore, Reason (R) is true.

Now, let's consider if Reason (R) is the correct explanation for Assertion (A).

Assertion (A) states that a circle is a special case of an ellipse when foci coincide. Reason (R) explains the geometric property that arises when the foci of an ellipse coincide (distance from center is constant) and notes that this property fits the definition of a circle. This property *is* what makes the figure a circle and connects it back to the ellipse by explaining what happens under the condition of coinciding foci. Thus, Reason (R) provides the justification for Assertion (A).

Reason (R) correctly explains Assertion (A).

Both Assertion (A) and Reason (R) are true, and Reason (R) is the correct explanation of Assertion (A).

The correct option is (A) Both A and R are true and R is the correct explanation of A.

The standard equation of a circle with center $(h, k)$ and radius $r$ is:

$(x - h)^2 + (y - k)^2 = r^2$

In this question, the center of the circle is given as $(0, 0)$. So, $h = 0$ and $k = 0$.

The radius is given as 7. So, $r = 7$.

Substitute the values of $h$, $k$, and $r$ into the standard equation:

$(x - 0)^2 + (y - 0)^2 = 7^2$

$x^2 + y^2 = 49$

This is the equation of the circle with center $(0, 0)$ and radius 7.

Let's examine the given options:

(A) $x^2 + y^2 = 7$: This represents a circle with center $(0,0)$ and radius $\sqrt{7}$, not 7.

(B) $x^2 + y^2 = 49$: This represents a circle with center $(0,0)$ and radius $\sqrt{49} = 7$. This matches our derived equation.

(C) $(x-7)^2 + (y-7)^2 = 0$: This equation is satisfied only when $x-7=0$ and $y-7=0$, i.e., $x=7$ and $y=7$. This represents a single point $(7, 7)$, which is a degenerate circle with radius 0.

(D) $x^2 + y^2 - 7 = 0$: This can be rewritten as $x^2 + y^2 = 7$, which is the same as option (A).

The equation of the circle is $x^2 + y^2 = 49$.

The correct option is (B) $x^2 + y^2 = 49$.

Given:

Vertex of the parabola is $(0, 0)$.

The axis of the parabola is along the x-axis.

The parabola passes through the point $(2, 3)$.

To Find:

The equation of the parabola.

Solution:

Since the vertex is at the origin $(0, 0)$ and the axis is along the x-axis, the standard equation of the parabola is of the form $y^2 = 4ax$ or $y^2 = -4ax$.

The standard equation for a parabola with vertex at the origin and axis along the x-axis is:

The parabola passes through the point $(2, 3)$. This means the coordinates of the point $(2, 3)$ must satisfy the equation of the parabola.

Substitute $x = 2$ and $y = 3$ into equation (i):

$3^2 = 4a(2)$

$9 = 8a$

Solve for $a$:

Now, substitute the value of $a$ back into the standard equation $y^2 = 4ax$ (equation (i)):

$y^2 = 4 \left(\frac{9}{8}\right) x$

$y^2 = \frac{36}{8} x$

Simplify the fraction $\frac{36}{8}$ by dividing both the numerator and the denominator by their greatest common divisor, which is 4.

$\frac{36}{8} = \frac{\cancel{36}^{9}}{\cancel{8}_{2}} = \frac{9}{2}$

So, the equation of the parabola is:

$y^2 = \frac{9}{2} x$

Comparing this equation with the given options, it matches option (A).

The correct option is (A) $y^2 = \frac{9}{2}x$.

Given the parametric equations:

From equation (i), we can write:

From equation (ii), we can write:

We use the fundamental trigonometric identity relating $\sec \theta$ and $\tan \theta$:

Substitute the expressions for $\sec \theta$ and $\tan \theta$ from equations (iii) and (iv) into equation (v):

$\left(\frac{x}{a}\right)^2 - \left(\frac{y}{b}\right)^2 = 1$

$\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$

This equation is the standard form of the equation of a hyperbola centered at the origin.

The standard form of a hyperbola with the transverse axis along the x-axis is $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$.

The standard form of a hyperbola with the transverse axis along the y-axis is $\frac{y^2}{a^2} - \frac{x^2}{b^2} = 1$.

The derived equation $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$ matches the standard form of a hyperbola with the transverse axis on the x-axis.

Therefore, the parametric equations $x = a \sec \theta, y = b \tan \theta$ represent a hyperbola.

The correct option is (D) A hyperbola.

The given equation of the hyperbola is $\frac{x^2}{16} - \frac{y^2}{9} = 1$.

This equation is in the standard form of a hyperbola centered at the origin with the transverse axis along the x-axis:

$\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$

Comparing the given equation with the standard form, we have:

$a^2 = 16$

$b^2 = 9$

Taking the square root of these values (and considering the positive roots for lengths):

$a = \sqrt{16} = 4$

$b = \sqrt{9} = 3$

For a hyperbola with the standard equation $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$:

The length of the transverse axis is $2a$.

The length of the conjugate axis is $2b$.

We are asked to find the length of the conjugate axis.

Length of the conjugate axis $= 2b = 2 \times 3 = 6$.

The length of the conjugate axis is 6.

Comparing this with the given options, it matches option (C).

The correct option is (C) 6.

Degenerate conic sections are the forms obtained when the intersecting plane passes through the vertex of the double-napped cone, or when the equation of a conic section represents a simpler geometric object.

Let's consider each option:

(A) A pair of intersecting lines: This is obtained when the plane passes through the vertex and intersects both nappes of the cone. An example is the equation $x^2 - y^2 = 0$, which factors into $(x-y)(x+y) = 0$, representing the two lines $y=x$ and $y=-x$. This is a degenerate hyperbola.

(B) A single point: This is obtained when the plane passes through the vertex and is perpendicular to the axis of the cone. An example is the equation $x^2 + y^2 = 0$, which is only satisfied by $(x, y) = (0, 0)$. This is a degenerate circle or ellipse.

(C) A pair of parallel lines: This can be obtained from a degenerate parabolic cylinder, or in some classifications as a degenerate parabola. An example is the equation $y^2 = k$ for $k > 0$, which gives $y = \pm \sqrt{k}$, representing two parallel horizontal lines. The equation $x^2 = k$ for $k > 0$ gives $x = \pm \sqrt{k}$, two parallel vertical lines. The equation $y^2 = 0$ gives $y=0$, which is a single line (also a degenerate parabola).

Since a pair of intersecting lines, a single point, and a pair of parallel lines are all considered degenerate conic sections, the correct answer is that all of the listed options are degenerate conic sections.

The correct option is (D) All of the above.

The given equation of the ellipse is $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$, with the condition $a > b$.

This is the standard form of an ellipse centered at the origin with the major axis along the x-axis.

Let's analyze each property listed in the options:

(A) Center is at the origin.

For the standard equation $\frac{(x-h)^2}{A^2} + \frac{(y-k)^2}{B^2} = 1$, the center is at $(h, k)$. In the given equation, $h=0$ and $k=0$.

So, the center is at $(0, 0)$, which is the origin.

This is a correct property.

(B) Foci are at $(\pm ae, 0)$.

For an ellipse with the major axis along the x-axis ($a > b$), the foci are located on the x-axis at a distance $c$ from the center, where $c^2 = a^2 - b^2$. The coordinates of the foci are $(\pm c, 0)$.

The eccentricity $e$ of an ellipse is defined as $e = \frac{c}{a}$. Thus, $c = ae$.

So, the foci are at $(\pm ae, 0)$.

This is a correct property.

(C) Vertices are at $(\pm a, 0)$.

For an ellipse with the major axis along the x-axis, the vertices are the endpoints of the major axis. They are located on the x-axis at a distance $a$ from the center $(0,0)$.

The coordinates of the vertices are $(\pm a, 0)$.

This is a correct property.

(D) The length of the latus rectum is $2a^2/b$.

The latus rectum of an ellipse is a chord passing through a focus and perpendicular to the major axis. The length of the latus rectum for an ellipse with the standard equation $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ is given by the formula $\frac{2b^2}{a}$.

The option states that the length of the latus rectum is $\frac{2a^2}{b}$. This formula is incorrect for this ellipse. The correct formula is $\frac{2b^2}{a}$.

This is NOT a correct property.

The question asks which of the following is NOT a property of the given ellipse.

Based on our analysis, option (D) is the property that is not correct.

The correct option is (D) The length of the latus rectum is $2a^2/b$.

The standard equation of a circle with center $(h, k)$ and radius $r$ is given by:

$(x - h)^2 + (y - k)^2 = r^2$

The given equation of the circle is:

$(x+5)^2 + (y-3)^2 = 25$

To find the center $(h, k)$, we compare the given equation with the standard form.

We can rewrite $(x+5)^2$ as $(x - (-5))^2$.

The equation becomes:

$(x - (-5))^2 + (y - 3)^2 = 25$

Comparing $(x - (-5))^2 + (y - 3)^2 = 25$ with $(x - h)^2 + (y - k)^2 = r^2$, we can identify the values of $h$, $k$, and $r^2$:

$h = -5$

$k = 3$

$r^2 = 25$

The center of the circle is $(h, k)$.

Center $= (-5, 3)$.

The radius is $r = \sqrt{25} = 5$.

Therefore, the center of the circle is $(-5, 3)$.

Comparing this result with the given options, it matches option (B).

The correct option is (B) $(-5, 3)$.

Given:

Equation of the line: $y = mx + c$

Equation of the circle: $x^2 + y^2 = r^2$

To Find:

The condition for the line to intersect the circle at two distinct points.

Solution:

The equation of the circle $x^2 + y^2 = r^2$ represents a circle centered at the origin $(0,0)$ with radius $r$.

A line intersects a circle at two distinct points if and only if the distance from the center of the circle to the line is less than the radius of the circle.

Let the center of the circle be $(x_0, y_0)$ and the radius be $R$. The equation of the line is $y = mx + c$. We can rewrite the line equation in the general form $Ax + By + C = 0$ as:

Here, $A=m$, $B=-1$, and $C=c$.

The center of the circle is $(x_0, y_0) = (0,0)$ and the radius is $R=r$.

The distance $d$ from a point $(x_0, y_0)$ to a line $Ax + By + C = 0$ is given by the formula:

$d = \frac{|Ax_0 + By_0 + C|}{\sqrt{A^2 + B^2}} $

Substitute the values of the center $(0,0)$ and the coefficients of the line $mx - y + c = 0$ into the distance formula:

$d = \frac{|m(0) + (-1)(0) + c|}{\sqrt{m^2 + (-1)^2}} $

$d = \frac{|0 + 0 + c|}{\sqrt{m^2 + 1}} $

$d = \frac{|c|}{\sqrt{m^2 + 1}} $

For the line to intersect the circle at two distinct points, the distance from the center to the line must be less than the radius:

$d < r $

Substitute the expression for $d$:

$\frac{|c|}{\sqrt{m^2 + 1}} < r $

Multiply both sides by $\sqrt{m^2 + 1}$ (since $\sqrt{m^2 + 1}$ is always positive):

$|c| < r \sqrt{m^2 + 1} $

or

$|c| < r \sqrt{1 + m^2} $

This is the condition for the line $y = mx + c$ to intersect the circle $x^2 + y^2 = r^2$ at two distinct points.

Let's consider the other cases for completeness:

If $|c| = r \sqrt{1 + m^2}$, the distance from the center to the line is equal to the radius, and the line is tangent to the circle (intersects at exactly one point).

If $|c| > r \sqrt{1 + m^2}$, the distance from the center to the line is greater than the radius, and the line does not intersect the circle.

Comparing the condition $|c| < r \sqrt{1 + m^2}$ with the given options, it matches option (A).

The correct option is (A) $|c| < r\sqrt{1+m^2}$.

The given equation of the parabola is $y^2 = 4ax$.

To determine the type of symmetry, we can test for symmetry about the x-axis, y-axis, and the origin.

Symmetry about the x-axis:

Replace $y$ with $-y$ in the equation. If the equation remains unchanged, the curve is symmetric about the x-axis.

Original equation: $y^2 = 4ax$

Substitute $y \to -y$: $(-y)^2 = 4ax$

Simplify: $y^2 = 4ax$

The equation remains the same. Thus, the parabola $y^2 = 4ax$ is symmetric about the x-axis.

Symmetry about the y-axis:

Replace $x$ with $-x$ in the equation. If the equation remains unchanged, the curve is symmetric about the y-axis.

Original equation: $y^2 = 4ax$

Substitute $x \to -x$: $y^2 = 4a(-x)$

Simplify: $y^2 = -4ax$

This is not the same as the original equation (unless $a=0$, which is a degenerate case). Thus, for $a \neq 0$, the parabola is not symmetric about the y-axis.

Symmetry about the origin:

Replace $x$ with $-x$ and $y$ with $-y$ in the equation. If the equation remains unchanged, the curve is symmetric about the origin.

Original equation: $y^2 = 4ax$

Substitute $x \to -x$ and $y \to -y$: $(-y)^2 = 4a(-x)$

Simplify: $y^2 = -4ax$

This is not the same as the original equation (unless $a=0$). Thus, for $a \neq 0$, the parabola is not symmetric about the origin.

The parabola $y^2 = 4ax$ is symmetric about the x-axis.

The correct option is (A) Symmetric about the x-axis.

Given:

Vertices of the ellipse are $(\pm 5, 0)$.

Foci of the ellipse are $(\pm 4, 0)$.

To Find:

The equation of the ellipse.

Solution:

The vertices $(\pm 5, 0)$ and foci $(\pm 4, 0)$ lie on the x-axis. This indicates that the major axis of the ellipse is along the x-axis.

The center of the ellipse is the midpoint of the vertices (or foci), which is $(\frac{5+(-5)}{2}, \frac{0+0}{2}) = (0, 0)$.

For an ellipse centered at the origin with the major axis along the x-axis, the standard equation is $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$, where $a$ is the length of the semi-major axis and $b$ is the length of the semi-minor axis.

The coordinates of the vertices are $(\pm a, 0)$. Comparing with the given vertices $(\pm 5, 0)$, we have:

The coordinates of the foci are $(\pm c, 0)$, where $c$ is the distance from the center to each focus. Comparing with the given foci $(\pm 4, 0)$, we have:

For an ellipse, the relationship between $a$, $b$, and $c$ is given by $c^2 = a^2 - b^2$ or $a^2 = b^2 + c^2$.

We need to find $b^2$. Using the relationship $b^2 = a^2 - c^2$:

$b^2 = 5^2 - 4^2$

$b^2 = 25 - 16$

Substitute the values of $a^2$ (from (i), $a^2=25$) and $b^2$ (from (iii), $b^2=9$) into the standard equation $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$:

$\frac{x^2}{25} + \frac{y^2}{9} = 1$

This is the equation of the ellipse.

Comparing this equation with the given options, it matches option (A).

The correct option is (A) $\frac{x^2}{25} + \frac{y^2}{9} = 1$.

The given equation of the hyperbola is $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$.

This is the standard form of a hyperbola centered at the origin with the transverse axis along the x-axis.

The equations of the asymptotes for a hyperbola with the standard form $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$ are given by the equation obtained by setting the right side to zero:

$\frac{x^2}{a^2} - \frac{y^2}{b^2} = 0$

Rearrange the equation to solve for $y$ in terms of $x$:

$\frac{y^2}{b^2} = \frac{x^2}{a^2}$

$y^2 = \frac{b^2}{a^2} x^2$

Take the square root of both sides:

$y = \pm \sqrt{\frac{b^2}{a^2} x^2}$

$y = \pm \frac{b}{a} \sqrt{x^2}$

$y = \pm \frac{b}{a} |x|$

For the lines themselves, we consider $y = \pm \frac{b}{a} x$.

The equations of the asymptotes are $y = \frac{b}{a}x$ and $y = -\frac{b}{a}x$, which can be written concisely as $y = \pm \frac{b}{a}x$.

Comparing this result with the given options, it matches option (A).

The correct option is (A) $y = \pm \frac{b}{a}x$.

Given:

The equation of the hyperbola is $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$.

To Find:

The parametric equation of the hyperbola.

Solution:

We need to find a pair of parametric equations for $x$ and $y$ in terms of a parameter (like $\theta$ or $t$) such that substituting them into the equation $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$ satisfies the equation.

Let's check each option:

(A) $x = a \cos \theta, y = b \sin \theta$

Substitute into the hyperbola equation:

$\frac{(a \cos \theta)^2}{a^2} - \frac{(b \sin \theta)^2}{b^2} = \frac{a^2 \cos^2 \theta}{a^2} - \frac{b^2 \sin^2 \theta}{b^2} = \cos^2 \theta - \sin^2 \theta$

This is equal to $\cos(2\theta)$, which is generally not equal to 1. These are the parametric equations for an ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$.

(B) $x = a t^2, y = 2 a t$

Substitute into the hyperbola equation:

$\frac{(a t^2)^2}{a^2} - \frac{(2 a t)^2}{b^2} = \frac{a^2 t^4}{a^2} - \frac{4 a^2 t^2}{b^2} = t^4 - \frac{4 a^2}{b^2} t^2$

This is generally not equal to 1. These are the parametric equations for a parabola $y^2 = 4ax$.

(C) $x = a \sec \theta, y = b \tan \theta$

Substitute into the hyperbola equation:

$\frac{(a \sec \theta)^2}{a^2} - \frac{(b \tan \theta)^2}{b^2} = \frac{a^2 \sec^2 \theta}{a^2} - \frac{b^2 \tan^2 \theta}{b^2} = \sec^2 \theta - \tan^2 \theta$

Using the trigonometric identity $\sec^2 \theta - \tan^2 \theta = 1$, we get:

$\sec^2 \theta - \tan^2 \theta = 1$

So, the equation $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$ is satisfied by these parametric equations.

(D) $x = r \cos \theta, y = r \sin \theta$

Substitute into the hyperbola equation:

$\frac{(r \cos \theta)^2}{a^2} - \frac{(r \sin \theta)^2}{b^2} = \frac{r^2 \cos^2 \theta}{a^2} - \frac{r^2 \sin^2 \theta}{b^2}$

This is generally not equal to 1 (unless $a=b=r$ and $\cos^2\theta - \sin^2\theta = 1$, which is not true for all $\theta$). These are the parametric equations for a circle $x^2 + y^2 = r^2$.

The parametric equations $x = a \sec \theta, y = b \tan \theta$ represent the hyperbola $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$.

The correct option is (C) $x = a \sec \theta, y = b \tan \theta$.

Given:

Center of the circle, $(h, k) = (0, 0)$.

The circle passes through the point $(x_1, y_1) = (3, 4)$.

To Find:

The equation of the circle.

Solution:

The standard equation of a circle with center $(h, k)$ and radius $r$ is:

$(x - h)^2 + (y - k)^2 = r^2$

Substitute the given center $(h, k) = (0, 0)$ into the standard equation:

$(x - 0)^2 + (y - 0)^2 = r^2$

The circle passes through the point $(3, 4)$. This means the coordinates of this point must satisfy the equation of the circle.

Substitute $x = 3$ and $y = 4$ into equation (i) to find the value of $r^2$:

$3^2 + 4^2 = r^2$

$9 + 16 = r^2$

$25 = r^2$

Substitute the value of $r^2$ from equation (ii) back into the equation (i):

$x^2 + y^2 = 25$

This is the equation of the circle with center $(0,0)$ and passing through the point $(3, 4)$.

Comparing this equation with the given options, it matches option (B).

The correct option is (B) $x^2 + y^2 = 25$.

The given equation of the parabola is $y^2 = -8x$.

This equation is in the standard form $y^2 = -4ax$, where the vertex is at the origin $(0, 0)$ and the axis of symmetry is along the x-axis. The negative sign indicates that the parabola opens to the left.

Comparing $y^2 = -8x$ with $y^2 = -4ax$, we have $-4a = -8$, which gives $a = 2$.

For a parabola with the equation $y^2 = kx$, the axis of symmetry is the x-axis.

The equation of the x-axis is $y = 0$.

Alternatively, the axis of symmetry is the line that passes through the vertex and the focus. The vertex is $(0,0)$. For $y^2 = -4ax$, the focus is at $(-a, 0)$. In this case, the focus is $(-2, 0)$. The line passing through $(0,0)$ and $(-2,0)$ is the x-axis, whose equation is $y=0$.

The equation of the axis of the parabola $y^2 = -8x$ is $y = 0$.

The correct option is (B) $y = 0$.

Given:

The equation of the hyperbola is $\frac{x^2}{16} - \frac{y^2}{9} = 1$.

To Find:

The foci of the hyperbola.

Solution:

The given equation is in the standard form of a hyperbola centered at the origin:

$\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$

Comparing the given equation $\frac{x^2}{16} - \frac{y^2}{9} = 1$ with the standard form, we have:

Since the $x^2$ term is positive, the transverse axis of the hyperbola is along the x-axis. The center of the hyperbola is at the origin $(0,0)$.

The foci of a hyperbola with transverse axis along the x-axis and centered at the origin are located at $(\pm c, 0)$, where the relationship between $a$, $b$, and $c$ is given by $c^2 = a^2 + b^2$.

Using the values from (i) and (ii), we calculate $c^2$:

$c^2 = a^2 + b^2$

$c^2 = 16 + 9$

$c^2 = 25$

Taking the square root of both sides to find $c$ (since $c$ represents a distance, we take the positive root):

$c = \sqrt{25}$

$c = 5$

The foci are located at $(\pm c, 0)$.

Substituting $c=5$, the coordinates of the foci are $(\pm 5, 0)$.

Comparing this result with the given options, it matches option (C).

The correct option is (C) $(\pm 5, 0)$.

The standard equation of a hyperbola centered at the origin depends on whether the transverse axis is along the x-axis or the y-axis.

If the transverse axis is along the x-axis, the standard equation is:

$\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$

In this case, the vertices are at $(\pm a, 0)$ and the foci are at $(\pm c, 0)$, where $c^2 = a^2 + b^2$.

If the transverse axis is along the y-axis, the standard equation is:

$\frac{y^2}{a^2} - \frac{x^2}{b^2} = 1$

In this case, the vertices are at $(0, \pm a)$ and the foci are at $(0, \pm c)$, where $c^2 = a^2 + b^2$.

Note that in both standard forms of the hyperbola, $a^2$ is always the denominator of the positive term, and $b^2$ is the denominator of the negative term. The length of the transverse axis is always $2a$, and the length of the conjugate axis is always $2b$.

The question asks for the equation of a hyperbola with the transverse axis along the y-axis and center at the origin.

This matches the second standard form mentioned above:

$\frac{y^2}{a^2} - \frac{x^2}{b^2} = 1$

Comparing this with the given options, it matches option (B).

The correct option is (B) $\frac{y^2}{a^2} - \frac{x^2}{b^2} = 1$.

Let's define the relationship between a line and a circle based on their points of intersection.

A line can intersect a circle in three ways:

1. The line does not intersect the circle at all.

2. The line intersects the circle at exactly one point.

3. The line intersects the circle at two distinct points.

Based on these possibilities, specific terms are used:

- If a line intersects a circle at two distinct points, it is called a secant.

- If a line intersects a circle at exactly one point, it is called a tangent.

- A chord is a line segment whose endpoints lie on the circle. A secant contains a chord.

- A diameter is a chord that passes through the center of the circle. It is part of a secant.

The question asks for the name of a line that intersects a circle at exactly one point.

This definition corresponds to a tangent line.

The correct option is (C) Tangent.

The given equation of the hyperbola is $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$.

This is the standard form of a hyperbola centered at the origin with the transverse axis along the x-axis.

In this form, $a$ is the length of the semi-transverse axis and $b$ is the length of the semi-conjugate axis.

The latus rectum of a hyperbola is a chord passing through a focus and perpendicular to the transverse axis.

For a hyperbola with the standard equation $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$, the length of each latus rectum is given by the formula $\frac{2b^2}{a}$.

Let's verify the formula. The foci are at $(\pm c, 0)$, where $c^2 = a^2 + b^2$. The equation of the line representing the latus rectum through the right focus is $x = c$. Substituting $x = c$ into the hyperbola equation:

$\frac{c^2}{a^2} - \frac{y^2}{b^2} = 1$

$\frac{y^2}{b^2} = \frac{c^2}{a^2} - 1 = \frac{c^2 - a^2}{a^2}$

Since $c^2 = a^2 + b^2$, we have $c^2 - a^2 = b^2$.

$\frac{y^2}{b^2} = \frac{b^2}{a^2}$

$y^2 = \frac{b^4}{a^2}$

$y = \pm \sqrt{\frac{b^4}{a^2}} = \pm \frac{b^2}{a}$

The endpoints of the latus rectum are $(c, \frac{b^2}{a})$ and $(c, -\frac{b^2}{a})$.

The length of the latus rectum is the distance between these two points:

Length $= |\frac{b^2}{a} - (-\frac{b^2}{a})| = |\frac{b^2}{a} + \frac{b^2}{a}| = |\frac{2b^2}{a}| $

Since $a$ and $b$ are lengths, they are positive, so the length is $\frac{2b^2}{a}$.

The length of the latus rectum of the hyperbola $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$ is $\frac{2b^2}{a}$.

Comparing this with the given options, it matches option (B).

The correct option is (B) $2b^2/a$.

Non-degenerate conic sections are the Circle, Ellipse, Parabola, and Hyperbola.

Let's analyze each option for all these conic sections.

(A) They all have a focus and a directrix.

A parabola is defined by one focus and one directrix ($e=1$).

An ellipse is defined by two foci and two corresponding directrices ($0 \leq e < 1$). Each point $P$ on the ellipse satisfies $PF_1 = e \cdot PD_1$ and $PF_2 = e \cdot PD_2$, where $D_1$ and $D_2$ are the feet of the perpendiculars from $P$ to the respective directrices.

A hyperbola is defined by two foci and two corresponding directrices ($e > 1$). Each point $P$ on the hyperbola satisfies $|PF_1 - PF_2| = \text{constant}$ and also $|PF_1| = e \cdot PD_1$ and $|PF_2| = e \cdot PD_2$.

A circle can be considered an ellipse with eccentricity $e=0$. While it can be defined using a focus (the center) and a directrix (a line at infinity), the standard definition of a conic section using a focus and a *finite* directrix typically applies when $e > 0$. Therefore, while true in a generalized sense, this statement is not universally presented as a defining property for a circle in the same way as for other conics with a finite directrix.

This statement is generally true for parabola, ellipse, and hyperbola. Its applicability to a circle depends on the specific definition used.

(B) They all have eccentricity greater than 0.

The eccentricity $e$ for the conic sections are:

• Circle: $e = 0$
• Ellipse: $0 \leq e < 1$
• Parabola: $e = 1$
• Hyperbola: $e > 1$

Since a circle is a non-degenerate conic section and has an eccentricity of 0, this statement is false.

(C) They are all symmetric with respect to at least one axis.

A circle is symmetric about any diameter passing through its center (infinitely many axes of symmetry).

An ellipse is symmetric about its major axis and its minor axis (two axes of symmetry).

A parabola is symmetric about its axis of symmetry (one axis of symmetry).

A hyperbola is symmetric about its transverse axis and its conjugate axis (two axes of symmetry).

In all cases, non-degenerate conic sections have at least one axis of symmetry.

This statement is true for all non-degenerate conic sections.

(D) All of the above are true.

Since statement (B) is false, this option is also false.

Based on the analysis, option (C) is the only property that is true for all non-degenerate conic sections.

The correct option is (C) They are all symmetric with respect to at least one axis.

The given equation is a general second-degree equation in two variables $x$ and $y$ with no $xy$ term:

$Ax^2 + By^2 + Cx + Dy + E = 0$

For a second-degree equation of the form $Ax^2 + Bxy + Cy^2 + Dx + Ey + F = 0$ to represent a non-degenerate conic section, we require that the discriminant $B^2 - 4AC$ (where $B$ is the coefficient of $xy$, $A$ is the coefficient of $x^2$, and $C$ is the coefficient of $y^2$ in the general form) is appropriate for the conic type.

In the given equation $Ax^2 + By^2 + Cx + Dy + E = 0$, the coefficient of $x^2$ is $A$, the coefficient of $y^2$ is $B$, and the coefficient of the $xy$ term is $0$. So, the discriminant $0^2 - 4AB = -4AB$.

For this equation to represent a circle, two main conditions must be met:

1. The coefficients of the $x^2$ and $y^2$ terms must be equal. So, $A = B$.

2. These equal coefficients must be non-zero. If $A=B=0$, the equation reduces to $Cx + Dy + E = 0$, which is the equation of a straight line (if $C$ or $D$ is non-zero).

Also, the discriminant condition for an ellipse (which includes a circle) is $B^2 - 4AC < 0$. Using the coefficients from the given equation, this is $0^2 - 4(A)(B) < 0$, which simplifies to $-4AB < 0$, or $AB > 0$. If $A=B$, then $AB = A^2$. So, $A^2 > 0$, which implies $A \neq 0$. Since $A=B$, this also means $B \neq 0$.

Combining these conditions, for the equation $Ax^2 + By^2 + Cx + Dy + E = 0$ to represent a circle, we must have $A = B$ and $A \neq 0$ (which implies $B \neq 0$).

Let's check the options:

(A) $A=B$ and $A \neq 0$: This condition implies $A=B \neq 0$ and $AB = A^2 > 0$, satisfying the requirements for a circle.

(B) $A=B=0$: This represents a linear equation (a line), not a circle.

(C) $A \neq 0, B \neq 0, A \neq B$: If $AB > 0$, this represents an ellipse. If $AB < 0$, this represents a hyperbola.

(D) $A=-B$: If $A = -B$ and $A, B \neq 0$, then $AB = A(-A) = -A^2$. Since $A \neq 0$, $-A^2 < 0$, so $AB < 0$. This condition represents a hyperbola.

The condition for the equation $Ax^2 + By^2 + Cx + Dy + E = 0$ to represent a circle is that $A$ and $B$ are equal and non-zero.

The correct option is (A) $A=B$ and $A \neq 0$.

Given:

Focus of the parabola, $F = (3, 0)$

Directrix of the parabola, $L$: $x = -3$

To Find:

The equation of the parabola.

Solution:

A parabola is defined as the locus of all points $(x, y)$ that are equidistant from a fixed point (the focus) and a fixed line (the directrix).

Let $P(x, y)$ be any point on the parabola.

The distance from $P(x, y)$ to the focus $F(3, 0)$ is $PF$.

$PF = \sqrt{(x - 3)^2 + (y - 0)^2} = \sqrt{(x - 3)^2 + y^2} $

The equation of the directrix is $x = -3$, which can be written as $x + 3 = 0$.

The distance from $P(x, y)$ to the directrix $L$ ($x + 3 = 0$) is $PD$.

Using the formula for the distance from a point $(x_1, y_1)$ to a line $Ax + By + C = 0$, which is $\frac{|Ax_1 + By_1 + C|}{\sqrt{A^2 + B^2}}$. For the point $(x, y)$ and the line $1x + 0y + 3 = 0$, the distance is:

$PD = \frac{|1 \cdot x + 0 \cdot y + 3|}{\sqrt{1^2 + 0^2}} = \frac{|x + 3|}{\sqrt{1}} = |x + 3| $

By the definition of a parabola, the distance from any point on the parabola to the focus is equal to the distance from that point to the directrix.

Substitute the expressions for $PF$ and $PD$ into equation (i):

$\sqrt{(x - 3)^2 + y^2} = |x + 3| $

To eliminate the square root and the absolute value, square both sides of the equation:

$(\sqrt{(x - 3)^2 + y^2})^2 = (|x + 3|)^2 $

$(x - 3)^2 + y^2 = (x + 3)^2 $

Expand the squared terms on both sides:

$(x^2 - 6x + 9) + y^2 = (x^2 + 6x + 9) $

Subtract $x^2$ and $9$ from both sides of the equation:

$-6x + y^2 = 6x $

Add $6x$ to both sides to isolate the $y^2$ term:

$y^2 = 6x + 6x $

This is the equation of the parabola.

Alternate Solution:

The focus is $(3, 0)$ and the directrix is $x = -3$.

Since the focus is on the x-axis and the directrix is a vertical line perpendicular to the x-axis, the axis of the parabola is the x-axis.

The vertex of the parabola is the midpoint between the focus $(3, 0)$ and the point on the directrix on the axis of symmetry $(-3, 0)$. The vertex is $(\frac{3 + (-3)}{2}, \frac{0 + 0}{2}) = (0, 0)$.

Since the vertex is at the origin $(0, 0)$ and the axis is along the x-axis, and the focus $(3,0)$ is on the positive x-axis, the standard equation of the parabola is $y^2 = 4ax$, where $a$ is the distance from the vertex to the focus (or vertex to directrix).

The distance from the vertex $(0, 0)$ to the focus $(3, 0)$ is $a = \sqrt{(3-0)^2 + (0-0)^2} = \sqrt{3^2} = 3$.

Substitute $a=3$ into the standard equation $y^2 = 4ax$:

$y^2 = 4(3)x $

$y^2 = 12x $

This confirms the result obtained using the definition.

Comparing the derived equation $y^2 = 12x$ with the given options, it matches option (A).

The correct option is (A) $y^2 = 12x$.

Given:

The equation of the ellipse is $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$, with the condition $a > b$.

To Find:

The distance between the vertices of the ellipse.

Solution:

The given equation $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ is the standard form of an ellipse centered at the origin $(0,0)$.

Since $a > b$, the major axis of the ellipse lies along the x-axis.

For an ellipse with the major axis along the x-axis and centered at the origin, the vertices are the endpoints of the major axis.

The coordinates of the vertices are $(\pm a, 0)$.

Let the two vertices be $V_1 = (a, 0)$ and $V_2 = (-a, 0)$.

The distance between these two vertices is the distance between the points $(a, 0)$ and $(-a, 0)$.

Using the distance formula $\sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$, the distance between $V_1$ and $V_2$ is:

Distance $= \sqrt{(-a - a)^2 + (0 - 0)^2} $

Distance $= \sqrt{(-2a)^2 + 0^2} $

Distance $= \sqrt{4a^2} $

Distance $= |2a| $

Since $a$ represents a semi-axis length, it is a positive value ($a>0$). Therefore, $|2a| = 2a$.

The distance between the vertices of the ellipse is $2a$. This distance is also known as the length of the major axis.

Comparing this result with the given options, it matches option (A).

The correct option is (A) $2a$.

The definition of a conic section based on the distance from fixed points (foci) is as follows:

A circle is the locus of points such that the distance from a single fixed point (the center) is constant.

A parabola is the locus of points such that the distance from a fixed point (the focus) is equal to the distance from a fixed line (the directrix).

An ellipse is the locus of points such that the sum of the distances from two fixed points (the foci) is constant.

A hyperbola is the locus of points such that the absolute difference of the distances from two fixed points (the foci) is constant.

The question states that the locus of a point moves such that the difference of its distances from two fixed points is constant.

This precisely matches the definition of a hyperbola.

The correct option is (D) Hyperbola.

Given:

The equation of the ellipse is $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$.

To Find:

The standard parametric equations for the ellipse.

Solution:

We need to find parametric equations for $x$ and $y$ in terms of a parameter (like $\theta$) such that substituting them into the equation $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ satisfies the equation.

Let's test the options by substituting the parametric forms for $x$ and $y$ into the ellipse equation.

(A) $x = a \cos \theta, y = b \sin \theta$

Substitute these into $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$:

Left side of the equation: $\frac{(a \cos \theta)^2}{a^2} + \frac{(b \sin \theta)^2}{b^2}$

$= \frac{a^2 \cos^2 \theta}{a^2} + \frac{b^2 \sin^2 \theta}{b^2} $

$= \cos^2 \theta + \sin^2 \theta $

Using the fundamental trigonometric identity $\cos^2 \theta + \sin^2 \theta = 1$, we get:

$= 1 $

This is equal to the right side of the ellipse equation ($\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$).

Thus, the parametric equations $x = a \cos \theta$ and $y = b \sin \theta$ satisfy the equation of the ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$.

Let's briefly look at the other options:

(B) $x = a \sec \theta, y = b \tan \theta$: Substituting into the hyperbola equation $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$ gives $\sec^2 \theta - \tan^2 \theta = 1$. These are parametric equations for a hyperbola.

(C) $x = a t^2, y = 2at$: Substituting into the parabola equation $y^2 = 4ax$ gives $(2at)^2 = 4a(at^2)$, which simplifies to $4a^2t^2 = 4a^2t^2$. These are parametric equations for a parabola.

(D) $x = r \cos \theta, y = r \sin \theta$: Substituting into the circle equation $x^2 + y^2 = r^2$ gives $(r \cos \theta)^2 + (r \sin \theta)^2 = r^2 (\cos^2 \theta + \sin^2 \theta) = r^2(1) = r^2$. These are parametric equations for a circle.

The parametric equations that represent the ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ are $x = a \cos \theta, y = b \sin \theta$.

The correct option is (A) $x = a \cos \theta, y = b \sin \theta$.

The eccentricity ($e$) is a parameter that determines the type of a conic section.

The eccentricity values for the non-degenerate conic sections are:

• Circle: The eccentricity is $e = 0$. (This is a special case of an ellipse).
• Ellipse: The eccentricity is $0 < e < 1$.
• Parabola: The eccentricity is $e = 1$.
• Hyperbola: The eccentricity is $e > 1$.

Now let's match the conic sections with the given eccentricity ranges:

(i) Circle $\implies e = 0$, which corresponds to option (d).

(ii) Ellipse $\implies 0 < e < 1$, which corresponds to option (c).

(iii) Parabola $\implies e = 1$, which corresponds to option (a).

(iv) Hyperbola $\implies e > 1$, which corresponds to option (b).

So, the correct matching is:

(i) - (d)

(ii) - (c)

(iii) - (a)

(iv) - (b)

Comparing this with the given options (A), (B), (C), (D):

(A) (i)-(d), (ii)-(c), (iii)-(a), (iv)-(b) - This matches our derived mapping.

(B) (i)-(c), (ii)-(d), (iii)-(a), (iv)-(b) - Incorrect.

(C) (i)-(d), (ii)-(c), (iii)-(b), (iv)-(a) - Incorrect.

(D) (i)-(d), (ii)-(b), (iii)-(a), (iv)-(c) - Incorrect.

The correct option is (A) (i)-(d), (ii)-(c), (iii)-(a), (iv)-(b).

Given:

The equation of the parabola is $y^2 = 4ax$ and the point on the parabola is $(x_1, y_1)$.

To Find:

The equation of the tangent to the parabola at the point $(x_1, y_1)$.

Solution:

The equation of the parabola is given by $y^2 = 4ax$.

To find the equation of the tangent line at a point $(x_1, y_1)$ on the parabola, we can use differentiation to find the slope at that point.

Differentiate the equation of the parabola with respect to $x$ implicitly:

$\frac{d}{dx}(y^2) = \frac{d}{dx}(4ax)$

$2y \frac{dy}{dx} = 4a$

Solving for $\frac{dy}{dx}$, we get the slope of the tangent at any point $(x, y)$ on the parabola:

$\frac{dy}{dx} = \frac{4a}{2y} = \frac{2a}{y}$

The slope of the tangent at the specific point $(x_1, y_1)$ is found by substituting $y_1$ for $y$:

$m = \left(\frac{dy}{dx}\right)_{(x_1, y_1)} = \frac{2a}{y_1}$

Now, we use the point-slope form of the equation of a line, which is $y - y_1 = m(x - x_1)$. Substitute the slope $m = \frac{2a}{y_1}$ and the point $(x_1, y_1)$:

$y - y_1 = \frac{2a}{y_1}(x - x_1)$

Multiply both sides of the equation by $y_1$ to clear the denominator:

$y_1(y - y_1) = 2a(x - x_1)$

Distribute on both sides:

$yy_1 - y_1^2 = 2ax - 2ax_1$

Since the point $(x_1, y_1)$ lies on the parabola $y^2 = 4ax$, it must satisfy the equation $y_1^2 = 4ax_1$. We can substitute this into the tangent equation:

$yy_1 - 4ax_1 = 2ax - 2ax_1$

Now, rearrange the terms to isolate $yy_1$ on one side:

$yy_1 = 2ax - 2ax_1 + 4ax_1$

$yy_1 = 2ax + 2ax_1$

Factor out $2a$ from the terms on the right side:

$yy_1 = 2a(x + x_1)$

This is the equation of the tangent to the parabola $y^2 = 4ax$ at the point $(x_1, y_1)$.

Comparing this result with the given options, we find that it matches option (A).

Alternate Method (using the standard formula derived from substitution):

For the parabola $y^2 = 4ax$, the equation of the tangent at a point $(x_1, y_1)$ on the parabola can be directly obtained by replacing $y^2$ with $yy_1$ and $2x$ (from $4ax = 2a \cdot 2x$) with $(x + x_1)$ in the equation of the parabola.

The equation $y^2 = 4ax$ becomes:

$yy_1 = 2a(x + x_1)$

This method provides the standard formula directly.

Both methods yield the same result.

The correct option is (A) $y y_1 = 2a(x + x_1)$.

Given:

The equation of the circle is $2x^2 + 2y^2 - 8x + 12y - 24 = 0$.

To Find:

The center and radius of the circle.

Solution:

The given equation of the circle is:

$2x^2 + 2y^2 - 8x + 12y - 24 = 0$

To find the center and radius, we convert this equation into the standard form of a circle equation, which is $(x-h)^2 + (y-k)^2 = r^2$, where $(h, k)$ is the center and $r$ is the radius.

First, divide the entire equation by the coefficient of $x^2$ and $y^2$, which is 2:

$\frac{2x^2 + 2y^2 - 8x + 12y - 24}{2} = \frac{0}{2}$

$x^2 + y^2 - 4x + 6y - 12 = 0$

Now, rearrange the terms, grouping the $x$ terms and $y$ terms together, and moving the constant term to the right side:

$(x^2 - 4x) + (y^2 + 6y) = 12$

We complete the square for the $x$ terms and the $y$ terms.

For the $x$ terms, take half of the coefficient of $x$ $(-4)$, square it $((-4/2)^2 = (-2)^2 = 4)$, and add it inside the parenthesis and to the right side:

$(x^2 - 4x + 4) + (y^2 + 6y) = 12 + 4$

$(x - 2)^2 + (y^2 + 6y) = 16$

For the $y$ terms, take half of the coefficient of $y$ $(6)$, square it $((6/2)^2 = 3^2 = 9)$, and add it inside the parenthesis and to the right side:

$(x - 2)^2 + (y^2 + 6y + 9) = 16 + 9$

$(x - 2)^2 + (y + 3)^2 = 25$

Rewrite the right side as a square:

Comparing equation (i) with the standard form of a circle equation $(x-h)^2 + (y-k)^2 = r^2$, we have:

Thus, the center of the circle is $(2, -3)$ and the radius is $5$.

Comparing this with the given options, we find that option (A) matches our result.

Alternate Method (using formulas for center and radius):

The general equation of a circle is $x^2 + y^2 + Dx + Ey + F = 0$.

The center $(h, k)$ is given by the formulas $h = -\frac{D}{2}$ and $k = -\frac{E}{2}$.

The radius $r$ is given by the formula $r = \sqrt{\left(\frac{D}{2}\right)^2 + \left(\frac{E}{2}\right)^2 - F} = \sqrt{h^2 + k^2 - F}$.

First, we must divide the given equation $2x^2 + 2y^2 - 8x + 12y - 24 = 0$ by 2 to get the form $x^2 + y^2 + Dx + Ey + F = 0$:

$x^2 + y^2 - 4x + 6y - 12 = 0$

Here, we have $D = -4$, $E = 6$, and $F = -12$.

Calculate the coordinates of the center $(h, k)$:

The center of the circle is $(2, -3)$.

Calculate the radius $r$:

$r = \sqrt{(2)^2 + (-3)^2 - (-12)}$

$r = \sqrt{4 + 9 + 12}$

$r = \sqrt{25}$

The radius of the circle is $5$.

Both methods confirm the center is $(2, -3)$ and the radius is 5.

The correct option is (A) Center $(2, -3)$, Radius 5.

Given:

The equation of the hyperbola is $\frac{y^2}{a^2} - \frac{x^2}{b^2} = 1$.

To Find:

The length of the latus rectum of the given hyperbola.

Solution:

The standard equation of a hyperbola with the transverse axis along the y-axis is given by:

$\frac{y^2}{a^2} - \frac{x^2}{b^2} = 1$

In this standard form, $a$ represents the length of the semi-transverse axis (along the y-axis) and $b$ represents the length of the semi-conjugate axis (along the x-axis).

The formula for the length of the latus rectum of a hyperbola is $\frac{2 \times (\text{length of semi-conjugate axis})^2}{\text{length of semi-transverse axis}}$.

For the hyperbola $\frac{y^2}{a^2} - \frac{x^2}{b^2} = 1$, the length of the semi-transverse axis is $a$ and the length of the semi-conjugate axis is $b$.

Therefore, the length of the latus rectum is:

Length of Latus Rectum $= \frac{2b^2}{a}$

Comparing this result with the given options, we find that it matches option (B).

The correct option is (B) $2b^2/a$.

Given:

Different forms of hyperbola equations are given as options.

To Find:

The equation that represents a rectangular hyperbola.

Solution:

A rectangular hyperbola (also called an equilateral hyperbola) is a hyperbola whose asymptotes are perpendicular to each other.

In the standard equation of a hyperbola centered at the origin, $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$ or $\frac{y^2}{a^2} - \frac{x^2}{b^2} = 1$, the lengths of the semi-transverse and semi-conjugate axes are $a$ and $b$ respectively.

A hyperbola is rectangular if and only if the lengths of its semi-axes are equal, i.e., $a=b$.

If $a=b$, the equation $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$ becomes $\frac{x^2}{a^2} - \frac{y^2}{a^2} = 1$, which simplifies to $x^2 - y^2 = a^2$.

This form matches option (A).

Another common form of a rectangular hyperbola, particularly one whose asymptotes are the coordinate axes, is $xy = c^2$ (where $c$ is a constant related to the size of the hyperbola).

This form matches option (C).

Option (B) $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$ with $a \neq b$ represents a general hyperbola, but not a rectangular one because the semi-axes lengths are not equal.

Since both $x^2 - y^2 = a^2$ and $xy = c^2$ are equations of rectangular hyperbolas (in different orientations), the correct option is the one that includes both possibilities.

Option (D) states "Both (A) and (C)".

Therefore, both $x^2 - y^2 = a^2$ and $xy = c^2$ represent rectangular hyperbolas.

The correct option is (D) Both (A) and (C).

Given:

The equation of the line is $y = mx + c$.

The equation of the circle is $x^2 + y^2 = r^2$.

To Find:

The condition for the line $y = mx + c$ to be tangent to the circle $x^2 + y^2 = r^2$.

Solution:

The standard equation of a circle centered at the origin $(0, 0)$ with radius $r$ is $x^2 + y^2 = r^2$.

The equation of the given line is $y = mx + c$.

We can rewrite the equation of the line in the general form $Ax + By + C = 0$ as $mx - y + c = 0$.

For a line to be tangent to a circle, the perpendicular distance from the center of the circle to the line must be equal to the radius of the circle.

The center of the circle $x^2 + y^2 = r^2$ is $(x_0, y_0) = (0, 0)$.

The radius of the circle is $r$.

The distance $d$ from a point $(x_0, y_0)$ to the line $Ax + By + C = 0$ is given by the formula:

$d = \frac{|Ax_0 + By_0 + C|}{\sqrt{A^2 + B^2}}$

Using the point $(0, 0)$ and the line $mx - y + c = 0$ (where $A=m$, $B=-1$, $C=c$), the distance $d$ from the center to the line is:

$d = \frac{|m(0) + (-1)(0) + c|}{\sqrt{m^2 + (-1)^2}}$

$d = \frac{|c|}{\sqrt{m^2 + 1}}$

For the line to be tangent to the circle, this distance $d$ must be equal to the radius $r$:

$d = r$

$\frac{|c|}{\sqrt{m^2 + 1}} = r$

To remove the absolute value and the square root, we square both sides of the equation:

$\left(\frac{|c|}{\sqrt{m^2 + 1}}\right)^2 = r^2$

$\frac{c^2}{m^2 + 1} = r^2$

Multiply both sides by $(m^2 + 1)$:

$c^2 = r^2 (m^2 + 1)$

Or, arranging the terms in the parenthesis:

$c^2 = r^2 (1+m^2)$

This is the required condition for the line $y = mx + c$ to be tangent to the circle $x^2 + y^2 = r^2$.

Comparing this result with the given options, we find that it matches option (A).

The correct option is (A) $c^2 = r^2 (1+m^2)$.

Given:

Focus of the parabola: $F = (2, 0)$

Vertex of the parabola: $V = (0, 0)$

To Find:

The equation of the parabola.

Solution:

The vertex of the parabola is at the origin $(0, 0)$.

The focus is given as $(2, 0)$.

Since the vertex is at $(0, 0)$ and the focus $(2, 0)$ lies on the positive x-axis, the axis of the parabola is the x-axis, and the parabola opens towards the positive x-direction.

The standard equation of a parabola with vertex at $(0, 0)$ and axis along the positive x-axis is of the form:

$y^2 = 4ax$

where $a$ is the distance from the vertex to the focus.

The focus of this standard parabola is at $(a, 0)$.

Comparing the given focus $(2, 0)$ with the standard focus $(a, 0)$, we have:

$a = 2$

Now, substitute the value of $a$ into the standard equation $y^2 = 4ax$:

$y^2 = 4(2)x$

This is the equation of the parabola.

Comparing this with the given options, we find that it matches option (B).

Alternate Solution (Using the Definition of a Parabola):

A parabola is the locus of a point that moves such that its distance from a fixed point (the focus) is equal to its distance from a fixed line (the directrix).

Given the vertex $V = (0, 0)$ and focus $F = (2, 0)$.

Since the vertex is the midpoint of the segment connecting the focus and the point where the axis intersects the directrix, and the focus is at $(2, 0)$, the directrix must be perpendicular to the x-axis and at a distance of 2 units to the left of the vertex.

The equation of the directrix is $x = -2$, or $x + 2 = 0$.

Let $P(x, y)$ be any point on the parabola.

The distance from $P$ to the focus $F(2, 0)$ is $PF = \sqrt{(x-2)^2 + (y-0)^2}$.

The distance from $P$ to the directrix $x + 2 = 0$ is the perpendicular distance, which is $PD = \frac{|x + 2|}{\sqrt{1^2 + 0^2}} = |x + 2|$.

According to the definition of a parabola, $PF = PD$.

$\sqrt{(x-2)^2 + y^2} = |x + 2|$

Square both sides to eliminate the square root and absolute value (since both sides are non-negative):

$(x-2)^2 + y^2 = (x + 2)^2$

Expand both sides:

$(x^2 - 4x + 4) + y^2 = x^2 + 4x + 4$

Subtract $x^2$ and 4 from both sides:

$-4x + y^2 = 4x$

Add $4x$ to both sides:

$y^2 = 4x + 4x$

Both methods yield the same equation for the parabola.

The correct option is (B) $y^2 = 8x$.

Given:

Equation of the ellipse: $\frac{x^2}{9} + \frac{y^2}{25} = 1$.

Assertion (A): The length of the major axis is 10.

Reason (R): The major axis is along the y-axis, and its length is $2a$, where $a^2 = 25$, so $a=5$. Length is $2 \times 5 = 10$.

To Find:

Whether Assertion (A) and Reason (R) are true and if Reason (R) is the correct explanation of Assertion (A).

Solution:

The given equation of the ellipse is $\frac{x^2}{9} + \frac{y^2}{25} = 1$.

This equation is in the standard form $\frac{x^2}{b^2} + \frac{y^2}{a^2} = 1$, where $a^2$ is the larger denominator.

Comparing the given equation with the standard form, we have:

$a^2 = 25$

$b^2 = 9$

Since $a^2 = 25$ and $b^2 = 9$, we have $a = \sqrt{25} = 5$ (assuming $a>0$) and $b = \sqrt{9} = 3$ (assuming $b>0$).

Because the larger denominator ($a^2 = 25$) is under the $y^2$ term, the major axis of the ellipse is along the y-axis.

The length of the major axis of an ellipse is $2a$.

Length of major axis $= 2 \times a = 2 \times 5 = 10$.

Let's evaluate the Assertion (A):

Assertion (A) states that the length of the major axis is 10. Our calculation shows the length is indeed 10.

Therefore, Assertion (A) is true.

Let's evaluate the Reason (R):

Reason (R) states: "The major axis is along the y-axis, and its length is $2a$, where $a^2 = 25$, so $a=5$. Length is $2 \times 5 = 10$."

Based on our analysis:

- "The major axis is along the y-axis": True, as the larger denominator is under $y^2$.

- "its length is $2a$": True, this is the formula for the length of the major axis.

- "where $a^2 = 25$": True, this is correctly identified from the equation.

- "so $a=5$": True, the square root of 25 is 5.

- "Length is $2 \times 5 = 10$": True calculation based on $a=5$.

Every part of Reason (R) is true. Therefore, Reason (R) is true.

Now, let's check if Reason (R) is the correct explanation for Assertion (A).

Reason (R) correctly identifies the orientation of the major axis, provides the correct formula for its length, extracts the correct value of $a$ from the given equation, and performs the correct calculation using this value of $a$ to arrive at the length of the major axis stated in Assertion (A).

Thus, Reason (R) provides a correct and complete explanation for why Assertion (A) is true.

Both Assertion (A) and Reason (R) are true, and Reason (R) is the correct explanation of Assertion (A).

This matches option (A).

The correct option is (A) Both A and R are true and R is the correct explanation of A.

Given:

The equation of the hyperbola is $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$.

The point on the hyperbola is $(x_1, y_1)$.

To Find:

The equation of the tangent to the hyperbola at the point $(x_1, y_1)$.

Solution:

The given equation of the hyperbola is:

$\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$

For a conic section given by a general second-degree equation $Ax^2 + Bxy + Cy^2 + Dx + Ey + F = 0$, the equation of the tangent at a point $(x_1, y_1)$ on the conic is obtained by the rule of replacing terms as follows:

$x^2 \to x x_1$

$y^2 \to y y_1$

$xy \to \frac{1}{2}(x y_1 + y x_1)$

$x \to \frac{1}{2}(x + x_1)$

$y \to \frac{1}{2}(y + y_1)$

Applying this rule to the equation of the hyperbola $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$ at the point $(x_1, y_1)$:

Replace $x^2$ with $x x_1$:

$\frac{x x_1}{a^2}$

Replace $y^2$ with $y y_1$:

$\frac{y y_1}{b^2}$

Substitute these replacements into the hyperbola equation:

$\frac{x x_1}{a^2} - \frac{y y_1}{b^2} = 1$

This is the equation of the tangent to the hyperbola $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$ at the point $(x_1, y_1)$.

Comparing this result with the given options, we find that it matches option (B).

The correct option is (B) $\frac{x x_1}{a^2} - \frac{y y_1}{b^2} = 1$.

Given:

The equation of the ellipse is $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ with the condition $a > b$.

To Find:

The equation of the directrix of the ellipse.

Solution:

The given equation of the ellipse is $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$.

The condition $a > b$ means that the denominator under the $x^2$ term ($a^2$) is greater than the denominator under the $y^2$ term ($b^2$).

This indicates that the major axis of the ellipse lies along the x-axis.

For an ellipse with the major axis along the x-axis ($\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$, where $a$ is the semi-major axis length and $b$ is the semi-minor axis length, and $a > b$), the foci are located at $(\pm ae, 0)$, where $e$ is the eccentricity.

The equations of the directrices for this type of ellipse are given by:

where $e$ is the eccentricity, calculated as $e = \sqrt{1 - \frac{b^2}{a^2}}$. Note that $0 < e < 1$ for an ellipse.

Comparing the formula for the directrices with the given options:

(A) $x = \pm ae$ represents the x-coordinates of the foci.

(B) $x = \pm a/e$ matches the standard form of the directrices when the major axis is along the x-axis.

(C) $y = \pm be$ is related to the y-coordinates of points on the ellipse or other properties, not the directrix for this orientation.

(D) $y = \pm b/e$ would be the directrix equation if the major axis were along the y-axis, and the equation was $\frac{x^2}{b^2} + \frac{y^2}{a^2} = 1$ with $a>b$ (where $a$ is under $y^2$).

Since the major axis is along the x-axis ($a>b$ in the denominator of $x^2$), the directrices are vertical lines with the equation $x = \pm \frac{a}{e}$.

The correct option is (B) $x = \pm a/e$.

Given:

Points on the circle: $(4, 1)$, $(6, 5)$.

Line containing the center $(h, k)$: $4h + k = 16$.

To Find:

Equation of the circle.

Solution:

Let the center be $(h, k)$. Since the center lies on $4x+y=16$, we have:

The distance from the center to $(4, 1)$ equals the distance from the center to $(6, 5)$ (both are the radius). Squaring these distances:

$(h-4)^2 + (k-1)^2 = (h-6)^2 + (k-5)^2$

$h^2 - 8h + 16 + k^2 - 2k + 1 = h^2 - 12h + 36 + k^2 - 10k + 25$

$-8h - 2k + 17 = -12h - 10k + 61$

$4h + 8k = 44$

Solve the system of equations (i) and (ii):

$4h + k = 16$

$h + 2k = 11$

From (i), $k = 16 - 4h$. Substitute into (ii):

$h + 2(16 - 4h) = 11$

$h + 32 - 8h = 11$

$-7h = -21 \implies h = 3$

Substitute $h=3$ into $k = 16 - 4h$: $k = 16 - 4(3) = 16 - 12 = 4$.

The center is $(3, 4)$.

Calculate the radius squared $r^2$ using the point $(4, 1)$ and center $(3, 4)$:

$r^2 = (4-3)^2 + (1-4)^2 = 1^2 + (-3)^2 = 1 + 9 = 10$

The equation of the circle with center $(3, 4)$ and $r^2 = 10$ is:

$(x - 3)^2 + (y - 4)^2 = 10$

$x^2 - 6x + 9 + y^2 - 8y + 16 = 10$

$x^2 + y^2 - 6x - 8y + 25 = 10$

$x^2 + y^2 - 6x - 8y + 15 = 0$

Comparing with the options, option (A) $x^2 + y^2 - 6x - 8y + 12 = 0$ has the correct coefficients for $x$ and $y$, implying the correct center $(3, 4)$. The constant term differs, suggesting a likely typo in the question or options. Based on the calculated center, option (A) is the closest match.

The correct option is likely (A) $x^2 + y^2 - 6x - 8y + 12 = 0$ (assuming a typo in the constant). The calculated correct equation is $x^2 + y^2 - 6x - 8y + 15 = 0$.

Given:

The equation of the ellipse is $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ with the condition $a > b$.

To Find:

The distance between the directrices of the ellipse.

Solution:

The given equation of the ellipse is $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$.

The condition $a > b$ indicates that the major axis of the ellipse lies along the x-axis. Here, $a$ is the length of the semi-major axis and $b$ is the length of the semi-minor axis.

For an ellipse with the major axis along the x-axis, centered at the origin, the equations of the directrices are given by:

where $e$ is the eccentricity of the ellipse.

The two directrices are the vertical lines $x = \frac{a}{e}$ and $x = -\frac{a}{e}$.

The distance between these two parallel vertical lines is the absolute difference of their x-coordinates:

Distance $= \left| \frac{a}{e} - \left(-\frac{a}{e}\right) \right|$

Distance $= \left| \frac{a}{e} + \frac{a}{e} \right|$

Distance $= \left| \frac{2a}{e} \right|$

Since $a > 0$ (length of semi-major axis) and $e > 0$ (eccentricity), $\frac{2a}{e}$ is positive.

Comparing this result with the given options, we find that it matches option (B).

The correct option is (B) $2a/e$.

Given:

The parametric equations of the point $(x, y)$ are $x = 2t^2$ and $y = 4t$, where $t$ is a parameter.

To Find:

The locus of the point $(x, y)$.

Solution:

We are given the parametric equations:

To find the locus, we need to eliminate the parameter $t$ from these two equations and obtain a relationship between $x$ and $y$.

From equation (ii), we can solve for $t$ in terms of $y$:

$t = \frac{y}{4}$

Now, substitute this expression for $t$ into equation (i):

$x = 2 \left(\frac{y}{4}\right)^2$

$x = 2 \left(\frac{y^2}{16}\right)$

$x = \frac{2y^2}{16}$

$x = \frac{y^2}{8}$

Rearrange the equation to isolate the squared term:

$y^2 = 8x$

This equation is of the form $y^2 = 4ax$, which is the standard equation of a parabola with its vertex at the origin $(0, 0)$ and its axis along the positive x-axis. The value of $a$ in this case is 2.

The equation $y^2 = 8x$ represents a parabola.

Therefore, the locus of the point $(x, y)$ defined by the given parametric equations is a parabola.

Comparing this result with the given options, we find that it matches option (B).

The correct option is (B) A parabola.

Given:

The equation of the ellipse is $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$.

The point on the ellipse is $(x_1, y_1)$.

To Find:

The equation of the tangent to the ellipse at the point $(x_1, y_1)$.

Solution:

The given equation of the ellipse is:

$\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$}

To find the equation of the tangent line at a point $(x_1, y_1)$ on a conic section, we can use the replacement rule for second-degree terms:

$x^2 \to x x_1$

$y^2 \to y y_1$

$xy \to \frac{1}{2}(x y_1 + y x_1)$

$x \to \frac{1}{2}(x + x_1)$

$y \to \frac{1}{2}(y + y_1)$

Applying this rule to the equation of the ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ at the point $(x_1, y_1)$:

Replace $x^2$ with $x x_1$:

$\frac{x x_1}{a^2}$

Replace $y^2$ with $y y_1$:

$\frac{y y_1}{b^2}$

Substitute these replacements into the ellipse equation:

$\frac{x x_1}{a^2} + \frac{y y_1}{b^2} = 1$}

This is the equation of the tangent to the ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ at the point $(x_1, y_1)$.

Comparing this result with the given options, we find that it matches option (B).

The correct option is (B) $\frac{x x_1}{a^2} + \frac{y y_1}{b^2} = 1$.

Given:

Equation of the hyperbola: $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$.

Points on the hyperbola: $(\pm 5, 0)$.

Distance between foci: 13 units.

To Find:

The value of $a$.

Solution:

The equation of the hyperbola is given in the standard form $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$.

This form indicates that the transverse axis is along the x-axis.

For a hyperbola with this orientation and centered at the origin, the vertices are located at $(\pm a, 0)$.

We are given that the points $(\pm 5, 0)$ are on the hyperbola.

Since these points lie on the x-axis and are symmetric about the origin, they must be the vertices of the hyperbola.

Comparing the coordinates of the given points $(\pm 5, 0)$ with the standard vertices $(\pm a, 0)$, we can identify the value of $a$.

Thus, the semi-transverse axis length $a$ is equal to $5$.

The information about the distance between the foci (13 units) is used to find the eccentricity $e$ or the value of $b$, but it is not necessary to find the value of $a$ itself, as $a$ is directly given by the location of the vertices $(\pm a, 0)$ on the transverse axis.

The distance between the foci is $2ae = 13$. Since $a=5$, $2(5)e = 13$, which gives $10e=13$, so $e = 13/10$. Using the relation $b^2 = a^2(e^2-1)$ for this hyperbola, we could find $b^2$, but the question only asks for $a$.

The value of $a$ is 5.

Comparing this result with the given options, we find that it matches option (A).

The correct option is (A) 5.

Given:

Equation of the hyperbola: $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$.

The hyperbola is centered at the origin $(0, 0)$.

Distance between the foci is 13 units.

To Find:

The distance from the center to each focus.

Solution:

For a hyperbola centered at the origin $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$, the foci are located at $(\pm c, 0)$, where $c$ is the distance from the center to each focus.

The distance between the two foci is $2c$.

We are given that the distance between the foci is 13 units.

So, $2c = 13$.}

To find the distance from the center (origin) to each focus, we need to find the value of $c$.

Divide the equation $2c = 13$ by 2:

$c = \frac{13}{2}$}

The distance from the center to each focus is $c = \frac{13}{2}$.

We can also express this as a decimal: $\frac{13}{2} = 6.5$.

Comparing this result with the given options:

(A) 13 is the distance between the two foci, not from the center to each focus.

(B) 13/2 matches our calculated distance.

(C) 6.5 is the decimal equivalent of 13/2 and also matches our calculated distance.

(D) Both (B) and (C) are correct as 13/2 and 6.5 are the same value.

The correct option is the one that includes both equivalent forms of the correct distance.

The correct option is (D) Both (B) and (C).

Given:

Equation of the hyperbola: $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$.

From Question 86, $a = 5$.

From Question 87, the distance from the center to each focus is $c = \frac{13}{2}$.

To Find:

The value of $b^2$.

Solution:

For a hyperbola with the standard equation $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$, centered at the origin, the relationship between the semi-transverse axis length $a$, the semi-conjugate axis length $b$, and the distance from the center to the focus $c$ is given by:

We need to find $b^2$. We can rearrange the formula to solve for $b^2$:

We are given $a = 5$ and $c = \frac{13}{2}$.

Substitute these values into equation (i):

$b^2 = \left(\frac{13}{2}\right)^2 - (5)^2$

Calculate the squares:

$\left(\frac{13}{2}\right)^2 = \frac{13^2}{2^2} = \frac{169}{4}$

$(5)^2 = 25$

Substitute these values back into the expression for $b^2$:

$b^2 = \frac{169}{4} - 25$

To subtract the numbers, find a common denominator, which is 4.

$25 = \frac{25 \times 4}{1 \times 4} = \frac{100}{4}$

Now subtract the fractions:

$b^2 = \frac{169}{4} - \frac{100}{4}$

$b^2 = \frac{169 - 100}{4}$

Comparing this result with the given options, we see that option (A) presents the same calculation steps and the final result.

The correct option is (A) $b^2 = (13/2)^2 - 5^2 = 169/4 - 25 = (169 - 100)/4 = 69/4$.

Given:

The general equation of the hyperbola is $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$.

From Question 86, the value of $a = 5$, which means $a^2 = 5^2 = 25$.

From Question 88, the value of $b^2 = \frac{69}{4}$.

To Find:

The equation of the hyperbola.

Solution:

The standard form of the equation of the hyperbola with the transverse axis along the x-axis and centered at the origin is $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$.

We have found the value of $a^2 = 25$ from the location of the vertices $(\pm 5, 0)$.

We have found the value of $b^2 = \frac{69}{4}$ using the distance between the foci.

Substitute these values of $a^2$ and $b^2$ into the standard equation of the hyperbola:

$\frac{x^2}{25} - \frac{y^2}{69/4} = 1$

This can also be written as:

$\frac{x^2}{25} - \frac{4y^2}{69} = 1$

Now, let's compare this equation with the given options:

(A) $\frac{x^2}{25} - \frac{y^2}{144} = 1$ - This would imply $b^2 = 144$, which is not what we found.

(B) $\frac{x^2}{25} - \frac{y^2}{69/4} = 1$ - This matches our calculated equation.

(C) $\frac{x^2}{144} - \frac{y^2}{25} = 1$ - This form implies the transverse axis is along the x-axis but with $a^2 = 144$, which means $a=12$. This contradicts the given vertices $(\pm 5, 0)$.

(D) $\frac{x^2}{25} + \frac{y^2}{144} = 1$ - This is the equation of an ellipse, not a hyperbola.

The equation of the hyperbola is $\frac{x^2}{25} - \frac{y^2}{69/4} = 1$.

Comparing with the options, option (B) is the correct one.

The correct option is (B) $\frac{x^2}{25} - \frac{y^2}{69/4} = 1$.

Given:

The equation of the parabola is $x^2 = 4ay$.

To Find:

The standard parametric equations for the parabola $x^2 = 4ay$.

Solution:

The standard parametric equations for a parabola are pairs of equations $(x, y)$ in terms of a parameter (usually $t$ or $\theta$) that satisfy the equation of the parabola for all values of the parameter.

Consider the given options and substitute them into the equation of the parabola $x^2 = 4ay$.

Option (A): $x = at^2, y = 2at$

Substitute these into $x^2 = 4ay$:

$(at^2)^2 = 4a(2at)$

$a^2 t^4 = 8a^2 t$

This equation is not true for all values of $t$. For example, if $t=1$, $a^2 = 8a^2$, which implies $a=0$ (if $a \neq 0$, this is false). So, this is not the correct parametric form.

Option (B): $x = 2at, y = at^2$

Substitute these into $x^2 = 4ay$:

$(2at)^2 = 4a(at^2)$

$4a^2 t^2 = 4a^2 t^2$

This equation is an identity and holds true for all values of the parameter $t$. Therefore, these are the parametric equations for the parabola $x^2 = 4ay$.

Option (C): $x = a \cos \theta, y = b \sin \theta$

These are the parametric equations for an ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$. They do not satisfy the equation of a parabola in general.

Option (D): $x = a \sec \theta, y = b \tan \theta$

These are related to the parametric equations for a hyperbola $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$. They do not satisfy the equation of a parabola in general.

Based on the substitution and verification, the standard parametric equations for the parabola $x^2 = 4ay$ are $x = 2at$ and $y = at^2$.

The correct option is (B) $x = 2at, y = at^2$.

Given:

Different mathematical expressions and descriptions related to conic sections.

To Find:

Which option represents a point conic.

Solution:

A point conic is a degenerate conic section. Degenerate conic sections occur when the intersecting plane passes through the apex of the double cone.

A point conic is essentially a single point.

Let's examine the given options:

(A) $x^2 + y^2 = 0$

This equation involves the sum of squares of real numbers. For the sum of squares to be zero, each individual term must be zero.

So, $x^2 = 0 \implies x = 0$, and $y^2 = 0 \implies y = 0$.

The only real solution to the equation $x^2 + y^2 = 0$ is the point $(0, 0)$. This is a single point.

Thus, the equation $x^2 + y^2 = 0$ represents a single point, which is a point conic (specifically, a degenerate circle).

(B) $x^2 + y^2 = -1$

The sum of squares of any real numbers $x$ and $y$ is always non-negative ($x^2 \geq 0$, $y^2 \geq 0$, so $x^2 + y^2 \geq 0$).

The equation $x^2 + y^2 = -1$ has no real solutions for $x$ and $y$. This equation represents an imaginary circle and does not correspond to a real geometric locus.

(C) A single point

By definition, a point conic is a single point. This option explicitly states this definition.

Based on our analysis, the equation $x^2 + y^2 = 0$ represents a single point (option A), and a single point is the definition of a point conic (option C).

Therefore, both option (A) and option (C) represent a point conic, one algebraically and one geometrically.

Option (D) includes both (A) and (C).

The correct option is (D) Both (A) and (C).

Given:

The equation of the ellipse is $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$, with the condition $b > a$.

A proposed formula for eccentricity squared: $e^2 = 1 - \frac{b^2}{a^2}$.

To Find:

Whether the proposed formula for eccentricity squared is correct for the given ellipse with $b > a$.

Solution:

The standard equation of an ellipse centered at the origin is $\frac{x^2}{L_1} + \frac{y^2}{L_2} = 1$.

The lengths of the semi-major axis and semi-minor axis are the square roots of the larger and smaller denominators, respectively.

The eccentricity $e$ is related to the semi-major axis (denoted by $A$) and the semi-minor axis (denoted by $B$) by the formula $e^2 = 1 - \frac{B^2}{A^2}$. Note that $A > B$ always for an ellipse (unless it's a circle, where $A=B$).

In the given equation $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$, the denominators are $a^2$ and $b^2$.

We are given the condition $b > a$, which implies $b^2 > a^2$.

Since $b^2$ is the larger denominator, the semi-major axis is $A = \sqrt{b^2} = b$. The major axis is along the y-axis.

The smaller denominator is $a^2$, so the semi-minor axis is $B = \sqrt{a^2} = a$.

Using the correct formula for eccentricity squared, $e^2 = 1 - \frac{B^2}{A^2}$:

$e^2 = 1 - \frac{a^2}{b^2}$}

The formula provided in the question is $e^2 = 1 - \frac{b^2}{a^2}$.

Comparing the correct formula ($e^2 = 1 - \frac{a^2}{b^2}$) with the provided formula ($e^2 = 1 - \frac{b^2}{a^2}$), we can see they are different because $a^2/b^2 \neq b^2/a^2$ since $a \neq b$ (given $b>a$).

Therefore, the proposed formula $e^2 = 1 - \frac{b^2}{a^2}$ is incorrect for the ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ when $b > a$.

Option (A) states "Yes", which is false.

Option (B) states "No ($e^2 = 1 - a^2/b^2$)", which correctly identifies that the proposed formula is incorrect and provides the correct formula for this case.

Option (C) discusses the case $a=b$ (a circle) where $e=0$, and both formulas yield $e^2=0$. However, the question specifies $b > a$, so this case is not relevant.

Option (D) discusses the case $a>b$. If $a>b$, then the major axis is along the x-axis, $A=a$, $B=b$, and the correct formula is $e^2 = 1 - b^2/a^2$. In this case, the provided formula happens to be correct. However, the question specifically gives the condition $b > a$.

For the given condition $b > a$, the proposed formula $e^2 = 1 - \frac{b^2}{a^2}$ is incorrect. The correct formula is $e^2 = 1 - \frac{a^2}{b^2}$.

The correct option is (B) No ($e^2 = 1 - a^2/b^2$).

Given:

Foci of the ellipse: $(0, \pm 3)$.

Vertices of the ellipse: $(0, \pm 5)$.

To Find:

The length of the major axis of the ellipse.

Solution:

The coordinates of the foci are $(0, \pm 3)$ and the vertices are $(0, \pm 5)$.

Since both the foci and vertices lie on the y-axis, the major axis of the ellipse is along the y-axis.

The center of the ellipse is the midpoint of the segment joining the foci (or the vertices), which is $(0, 0)$.

For an ellipse centered at the origin with the major axis along the y-axis, the standard equation is $\frac{x^2}{b^2} + \frac{y^2}{a^2} = 1$, where $a > b$.

The vertices of this ellipse are at $(0, \pm a)$, and the foci are at $(0, \pm c)$, where $c$ is the distance from the center to the foci.

From the given vertices $(0, \pm 5)$, we have the length of the semi-major axis:

From the given foci $(0, \pm 3)$, we have the distance from the center to the foci:

The length of the major axis of the ellipse is given by $2a$.

Length of major axis $= 2 \times a$

Substitute the value of $a=5$:

Length of major axis $= 2 \times 5 = 10$

The length of the major axis is 10 units.

Comparing this result with the given options, we find that it matches option (C).

The correct option is (C) 10.

Given:

Description of a locus: The set of all points $(x, y)$ that are equidistant from a fixed point and a fixed line.

Fixed point: Called the focus ($F$).

Fixed line: Called the directrix ($D$).

To Find:

The name of the conic section represented by this locus.

Solution:

The definition of a conic section based on its eccentricity $e$ relates the distance from a point on the conic to the focus ($PF$) and the distance from the same point to the directrix ($PD$).

The relationship is given by $PF = e \cdot PD$.

For a point to be equidistant from the fixed point (focus) and the fixed line (directrix), the ratio of the distance from the point to the focus and the distance from the point to the directrix must be equal to 1.

This means $PF = 1 \cdot PD$, or $PF = PD$.

The value of the eccentricity $e$ determines the type of conic section:

• If $e = 1$, the conic is a parabola.
• If $0 \leq e < 1$, the conic is an ellipse.
• If $e > 1$, the conic is a hyperbola.
• If $e = 0$, the conic is a circle (where the directrix is at infinity).

In the given description, the distance from the point to the fixed point (focus) is equal to the distance from the point to the fixed line (directrix). This corresponds to the condition $PF = PD$, which implies the eccentricity $e = 1$.

A conic section with eccentricity $e = 1$ is a parabola.

Therefore, the locus of a point equidistant from a fixed point and a fixed line is a parabola.

Comparing this result with the given options, we find that it matches option (B).

The correct option is (B) Parabola.

Given:

Equation of the circle: $x^2 + y^2 - 2x - 4y - 4 = 0$.

Equation of a line: $x - y + 1 = 0$.

The required line passes through the center of the circle and is perpendicular to the given line.

To Find:

The equation of the required line.

Solution:

First, find the center of the circle $x^2 + y^2 - 2x - 4y - 4 = 0$.

The general equation of a circle is $x^2 + y^2 + Dx + Ey + F = 0$.

The coordinates of the center are $(h, k) = \left(-\frac{D}{2}, -\frac{E}{2}\right)$.

Comparing the given equation with the general form, we have $D = -2$ and $E = -4$.

Center $(h, k) = \left(-\frac{-2}{2}, -\frac{-4}{2}\right) = \left(\frac{2}{2}, \frac{4}{2}\right) = (1, 2)$.

So, the required line passes through the point $(1, 2)$.

Next, find the slope of the given line $x - y + 1 = 0$.

Rewrite the equation in the form $y = mx + c$:

$y = x + 1$

The slope of this line is $m_1 = 1$.

The required line is perpendicular to the line $x - y + 1 = 0$.

If two lines are perpendicular, the product of their slopes is $-1$ (assuming neither is vertical or horizontal).

Let the slope of the required line be $m_2$.

$m_1 \cdot m_2 = -1$

$1 \cdot m_2 = -1$

$m_2 = -1$

Now, we have the point $(1, 2)$ that the required line passes through and its slope $m_2 = -1$.

Use the point-slope form of the equation of a line: $y - y_1 = m(x - x_1)$.

$y - 2 = -1(x - 1)$

$y - 2 = -x + 1$

Rearrange the terms to get the equation in the form $Ax + By + C = 0$:

$x + y - 2 - 1 = 0$

$x + y - 3 = 0$

This is the equation of the line passing through the center of the circle and perpendicular to the given line.

Comparing this derived equation with the given options:

(A) $x+y-3=0$

(B) $x-y+1=0$

(C) $x+y+3=0$

(D) $x-y-1=0$

Our derived equation $x + y - 3 = 0$ matches option (A).

The correct option is (A) $x+y-3=0$.

Given:

The equation of the hyperbola is $\frac{y^2}{a^2} - \frac{x^2}{b^2} = 1$.

To Find:

The length of the transverse axis of the hyperbola.

Solution:

The given equation of the hyperbola is in the standard form where the term with $y^2$ is positive:

$\frac{y^2}{a^2} - \frac{x^2}{b^2} = 1$}

In this form, the transverse axis of the hyperbola lies along the y-axis.

The vertices of this hyperbola are located at $(0, \pm a)$.

The transverse axis is the segment connecting the two vertices. Its endpoints are $(0, -a)$ and $(0, a)$.

The length of the transverse axis is the distance between these two points:

Length of transverse axis $= \sqrt{(0-0)^2 + (a - (-a))^2}$

Length of transverse axis $= \sqrt{0^2 + (a+a)^2}$

Length of transverse axis $= \sqrt{(2a)^2}$

Length of transverse axis $= |2a|$

Since $a$ is usually taken as a positive value representing a length (semi-transverse axis), the length of the transverse axis is $2a$.

For a hyperbola $\frac{y^2}{a^2} - \frac{x^2}{b^2} = 1$, the denominator under the positive term ($y^2/a^2$) determines the value $a$, which is the length of the semi-transverse axis. The length of the transverse axis is $2a$.

If the equation were $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$, the transverse axis would be along the x-axis, vertices at $(\pm a, 0)$, and its length would still be $2a$. The notation implies $a$ is the semi-transverse axis length corresponding to the variable with the positive squared term.

In the given equation $\frac{y^2}{a^2} - \frac{x^2}{b^2} = 1$, $a^2$ is the denominator of the $y^2$ term. Thus, $a$ is the semi-transverse axis length.

The length of the transverse axis is $2a$.

Comparing this result with the given options, we find that it matches option (A).

The correct option is (A) $2a$.

Given:

The equation of the parabola is $x^2 = 4ay$.

To Find:

The equation of the directrix of the parabola.

Solution:

The standard equation of a parabola is determined by its orientation and vertex.

The equation $x^2 = 4ay$ represents a parabola with its vertex at the origin $(0, 0)$.

Since the $x$ term is squared, the axis of symmetry is along the y-axis.

If $a > 0$, the parabola opens upwards.

If $a < 0$, the parabola opens downwards.

For the standard parabola $x^2 = 4ay$, the focus is located at $(0, a)$ on the axis of symmetry.

The directrix of a parabola is a line perpendicular to the axis of symmetry.

It is located at a distance equal to the focal length ($|a|$) from the vertex, but on the opposite side of the focus.

Since the axis of symmetry is the y-axis (a vertical line), the directrix is a horizontal line.

The distance from the vertex $(0, 0)$ to the focus $(0, a)$ is $|a|$.

The directrix is on the opposite side of the vertex from the focus, along the y-axis.

Therefore, the directrix is a horizontal line with the equation $y = -a$.}

This line is perpendicular to the y-axis and is at a distance $|-a| = |a|$ from the vertex $(0, 0)$.

Comparing this result with the given options, we find that it matches option (D).

The correct option is (D) $y = -a$.

Given:

The length of the latus rectum of the ellipse is half the length of its minor axis.

To Find:

The eccentricity ($e$) of the ellipse.

Solution:

Let the semi-major axis length be $A$ and the semi-minor axis length be $B$. For an ellipse, $A > B$ (or $A=B$ for a circle).

The length of the minor axis is $2B$.

The length of the latus rectum of an ellipse is given by $\frac{2B^2}{A}$.

According to the problem statement:

Length of Latus Rectum $= \frac{1}{2} \times$ Length of Minor Axis

$\frac{2B^2}{A} = \frac{1}{2} (2B)$

$\frac{2B^2}{A} = B$

Since $B$ is the semi-minor axis length of a non-degenerate ellipse, $B > 0$. We can divide both sides by $B$:

$\frac{2B}{A} = 1$

$\displaystyle A = 2B$

This tells us that the length of the semi-major axis is twice the length of the semi-minor axis.

The eccentricity $e$ of an ellipse is related to the semi-major axis $A$ and semi-minor axis $B$ by the formula $B^2 = A^2 (1 - e^2)$.

We can rearrange this formula to solve for $e^2$:

$1 - e^2 = \frac{B^2}{A^2}$

$e^2 = 1 - \frac{B^2}{A^2}$

Substitute the relationship $A = 2B$ into this equation:

$e^2 = 1 - \frac{B^2}{(2B)^2}$

$e^2 = 1 - \frac{B^2}{4B^2}$

Since $B \neq 0$, we can cancel $B^2$:

$e^2 = 1 - \frac{1}{4}$

$e^2 = \frac{4}{4} - \frac{1}{4}$

$e^2 = \frac{3}{4}$

Taking the square root of both sides (and considering that eccentricity $e$ is non-negative for an ellipse):

$e = \sqrt{\frac{3}{4}}$

$e = \frac{\sqrt{3}}{\sqrt{4}}$

$\displaystyle e = \frac{\sqrt{3}}{2}$

The eccentricity of the ellipse is $\frac{\sqrt{3}}{2}$.

Comparing this result with the given options, we find that it matches option (C).

The correct option is (C) $\sqrt{3}/2$.

Given:

The equation of the parabola is $x^2 = 8y$.

To Find:

The length of the latus rectum of the parabola.

Solution:

The given equation of the parabola is $x^2 = 8y$.

This equation is in the standard form $x^2 = 4ay$, which represents a parabola with its vertex at the origin $(0, 0)$ and its axis along the y-axis.

Comparing the given equation $x^2 = 8y$ with the standard form $x^2 = 4ay$, we can equate the coefficients of $y$:

Solving for $a$:

$a = \frac{8}{4}$

For a parabola in the standard form $x^2 = 4ay$ or $y^2 = 4ax$, the length of the latus rectum is given by $|4a|$.

Using the value of $a$ from (i):

Length of Latus Rectum $= |4 \times 2|$

Length of Latus Rectum $= |8|$

The length of the latus rectum of the parabola $x^2 = 8y$ is 8 units.

Comparing this result with the given options, we find that it matches option (C).

The correct option is (C) 8.

Given:

The general second-degree equation $Ax^2 + Bxy + Cy^2 + Dx + Ey + F = 0$.

To Find:

The condition(s) under which the equation represents a circle.

Solution:

The type of conic section represented by the general second-degree equation $Ax^2 + Bxy + Cy^2 + Dx + Ey + F = 0$ is determined by the discriminant $B^2 - 4AC$.

A circle is a special case of an ellipse.

For the equation to represent a circle, the following conditions must be met:

1. The $xy$ term must be zero. This means the coefficient $B$ must be $0$.

2. The coefficients of the $x^2$ term ($A$) and the $y^2$ term ($C$) must be equal and non-zero ($A = C \neq 0$). If $A=C=0$, the equation reduces to a linear equation, which represents a line (a degenerate case but not a typical circle representation). If $A=C \neq 0$, we can divide the entire equation by $A$ to get $x^2 + y^2 + (D/A)x + (E/A)y + (F/A) = 0$, which is the standard general form of a circle equation.

Let's examine the discriminant $B^2 - 4AC$ under these conditions:

If $B=0$ and $A=C$, the discriminant becomes $0^2 - 4A \cdot A = -4A^2$.

For a real circle, $A \neq 0$. In this case, $-4A^2$ is always negative (since $A^2$ is positive). Thus, the condition $B^2 - 4AC < 0$ is implicitly satisfied when $B=0$ and $A=C \neq 0$.

So, the conditions $B=0$ and $A=C \neq 0$ are necessary and sufficient for the equation to represent a circle (or a point or no real locus, depending on the value of $F$, which are degenerate circles). Including $B^2 - 4AC < 0$ confirms it's an ellipse, and $A=C, B=0$ narrows it down to a circle.

Let's evaluate the options:

(A) $B^2 - 4AC = 0$: This is the condition for a parabola (or degenerate parabola). Incorrect.

(B) $B^2 - 4AC < 0$ and $A=C, B=0$:

- $B^2 - 4AC < 0$ indicates an ellipse (or degenerate forms).

- $A=C$ and $B=0$ indicate that the coefficients of $x^2$ and $y^2$ are equal and there is no $xy$ term.

These conditions together precisely describe a circle or a degenerate circle (point or no locus).

(C) $A \neq 0, B \neq 0, A=C$: If $B \neq 0$, there is an $xy$ term. This is not the form of a standard circle aligned with the axes. Incorrect.

(D) $A=-C$: If $A=-C$ and $B=0$ (i.e., $Ax^2 - Ay^2 + Dx + Ey + F = 0$), this represents a hyperbola (as $B^2 - 4AC = 0 - 4(A)(-A) = 4A^2 > 0$ if $A \neq 0$). Incorrect.

The condition for the equation $Ax^2 + Bxy + Cy^2 + Dx + Ey + F = 0$ to represent a circle is that the $xy$ term is absent ($B=0$) and the coefficients of $x^2$ and $y^2$ are equal and non-zero ($A=C \neq 0$). Option (B) includes $B^2 - 4AC < 0$ which is implied by $B=0, A=C \neq 0$, along with the explicit conditions $A=C$ and $B=0$. This correctly identifies the requirements.

The correct option is (B) $B^2 - 4AC < 0$ and $A=C, B=0$.

Given:

A description of how a conic section is formed by intersecting a double cone with a plane.

The intersecting plane is parallel to a generator (a line on the surface of the cone passing through the vertex).

To Find:

The type of conic section formed by this intersection.

Solution:

Conic sections are the curves formed by the intersection of a plane with a double cone. The type of conic section formed depends on the angle of the intersecting plane relative to the axis of the cone and the generators of the cone.

Let $\alpha$ be the angle between the axis of the cone and a generator (i.e., the semi-vertical angle of the cone).

Let $\beta$ be the angle between the intersecting plane and the axis of the cone.

The different conic sections are formed as follows:

• Circle: When the plane is perpendicular to the axis of the cone ($\beta = 90^\circ$). The plane intersects only one nappe of the cone.
• Ellipse: When the plane is tilted but not perpendicular to the axis, and it intersects only one nappe, and the angle $\beta$ is greater than the semi-vertical angle $\alpha$ ($\alpha < \beta < 90^\circ$).
• Parabola: When the plane is parallel to a generator of the cone ($\beta = \alpha$). The plane intersects only one nappe and is parallel to the slant edge of the cone.
• Hyperbola: When the plane is parallel to the axis of the cone ($\beta = 0^\circ$) or when the angle $\beta$ is less than the semi-vertical angle $\alpha$ ($0 \leq \beta < \alpha$). The plane intersects both nappes of the cone.

The question states that the intersecting plane is parallel to the generator of the cone.

This corresponds to the case where the angle $\beta$ between the plane and the axis of the cone is equal to the semi-vertical angle $\alpha$ of the cone ($\beta = \alpha$).

According to the definitions above, when $\beta = \alpha$, the conic section formed is a parabola.

Therefore, the locus of the points of intersection is a parabola.

Comparing this result with the given options, we find that it matches option (C).

The correct option is (C) Parabola.

The standard equation of a circle with center $(h, k)$ and radius $r$ is given by:

$(x - h)^2 + (y - k)^2 = r^2$

In this problem, the given center is $(h, k) = (2, -3)$ and the radius is $r = 4$.

Substituting these values into the standard equation, we get:

$(x - 2)^2 + (y - (-3))^2 = 4^2$

$(x - 2)^2 + (y + 3)^2 = 16$

Expanding the equation:

$(x^2 - 4x + 4) + (y^2 + 6y + 9) = 16$

$x^2 - 4x + 4 + y^2 + 6y + 9 = 16$

$x^2 + y^2 - 4x + 6y + 13 = 16$

$x^2 + y^2 - 4x + 6y + 13 - 16 = 0$

$x^2 + y^2 - 4x + 6y - 3 = 0$

Therefore, the equation of the circle is:

$\mathbf{(x - 2)^2 + (y + 3)^2 = 16}$

or in general form:

$\mathbf{x^2 + y^2 - 4x + 6y - 3 = 0}$

Given:

The equation of the circle is $x^2 + y^2 - 4x + 6y - 5 = 0$.

To Find:

The centre and radius of the given circle.

Solution:

We are given the equation of the circle in the general form: $x^2 + y^2 - 4x + 6y - 5 = 0$.

To find the centre and radius, we will convert this equation into the standard form $(x - h)^2 + (y - k)^2 = r^2$ by using the method of completing the square.

Group the x-terms and y-terms and move the constant term to the right side:

$(x^2 - 4x) + (y^2 + 6y) = 5$

Complete the square for the x-terms. The coefficient of $x$ is -4. Half of -4 is -2, and $(-2)^2 = 4$. Add 4 to both sides:

$(x^2 - 4x + 4) + (y^2 + 6y) = 5 + 4$

Complete the square for the y-terms. The coefficient of $y$ is 6. Half of 6 is 3, and $3^2 = 9$. Add 9 to both sides:

$(x^2 - 4x + 4) + (y^2 + 6y + 9) = 5 + 4 + 9$

Rewrite the expressions in parentheses as perfect squares and simplify the right side:

$(x - 2)^2 + (y + 3)^2 = 18$

This equation is now in the standard form $(x - h)^2 + (y - k)^2 = r^2$, where $(h, k)$ is the centre and $r$ is the radius.

Comparing $(x - 2)^2 + (y + 3)^2 = 18$ with the standard form, we have:

$h = 2$

$k = -3$

$r^2 = 18$

From these values, the centre of the circle is $(h, k) = \mathbf{(2, -3)}$.

The radius of the circle is $r = \sqrt{18}$.

We can simplify the radius:

$r = \sqrt{9 \times 2} = \sqrt{9} \times \sqrt{2} = 3\sqrt{2}$

So, the radius is $\mathbf{3\sqrt{2}}$.

Answer:

The centre of the circle is $\mathbf{(2, -3)}$ and the radius is $\mathbf{3\sqrt{2}}$.

Given:

Vertex of the parabola is at the origin $(0, 0)$.

Focus of the parabola is at $(0, 2)$.

To Find:

The equation of the parabola.

Solution:

The vertex of the parabola is at the origin $(0, 0)$.

The focus is at $(0, 2)$.

Since the vertex is at the origin and the focus is on the positive y-axis, the parabola opens upwards.

The standard equation of a parabola with vertex at the origin $(0, 0)$ and focus at $(0, a)$ opening upwards is given by:

$x^2 = 4ay$

Comparing the given focus $(0, 2)$ with the standard focus $(0, a)$, we find the value of $a$:

Thus, $a = 2$.

Substitute the value of $a$ into the standard equation $x^2 = 4ay$:

$x^2 = 4(2)y$

$x^2 = 8y$

The equation of the parabola with vertex at the origin and focus at $(0, 2)$ is $\mathbf{x^2 = 8y}$.

Given:

Foci of the ellipse are $(\pm 5, 0)$.

Vertices of the ellipse are $(\pm 13, 0)$.

To Find:

The equation of the ellipse.

Solution:

The foci are given as $(\pm 5, 0)$ and the vertices are given as $(\pm 13, 0)$.

Since the foci and vertices are on the x-axis and are symmetric with respect to the origin, the center of the ellipse is at the origin $(0, 0)$.

The standard equation of an ellipse centered at the origin is $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ for a horizontal ellipse or $\frac{x^2}{b^2} + \frac{y^2}{a^2} = 1$ for a vertical ellipse.

The distance from the center to a focus is denoted by $c$. From the foci $(\pm 5, 0)$, we have $c = 5$.

The distance from the center to a vertex is denoted by $a$. From the vertices $(\pm 13, 0)$, we have $a = 13$.

Since the vertices are on the x-axis $(\pm a, 0)$, this is a horizontal ellipse.

The standard equation for a horizontal ellipse centered at the origin is:

$\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$

We know $a = 13$, so $a^2 = 13^2 = 169$.

We know $c = 5$, so $c^2 = 5^2 = 25$.

For an ellipse, the relationship between $a$, $b$, and $c$ is $c^2 = a^2 - b^2$.

We can use this to find $b^2$:

$b^2 = a^2 - c^2$

$b^2 = 169 - 25$

$b^2 = 144$

Now, substitute the values of $a^2 = 169$ and $b^2 = 144$ into the standard equation of the horizontal ellipse:

$\frac{x^2}{169} + \frac{y^2}{144} = 1$

The equation of the ellipse is $\mathbf{\frac{x^2}{169} + \frac{y^2}{144} = 1}$.

Given:

Foci of the hyperbola are $(\pm \sqrt{5}, 0)$.

Length of the transverse axis is 4.

To Find:

The equation of the hyperbola.

Solution:

The foci are given as $(\pm \sqrt{5}, 0)$. Since the foci are on the x-axis and are symmetric with respect to the origin, the center of the hyperbola is at the origin $(0, 0)$.

The standard equation of a hyperbola centered at the origin with foci on the x-axis (transverse axis horizontal) is:

$\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$

The distance from the center to a focus is denoted by $c$. From the foci $(\pm \sqrt{5}, 0)$, we have $c = \sqrt{5}$.

Squaring both sides, $c^2 = (\sqrt{5})^2 = 5$.

The length of the transverse axis is given as 4. For a horizontal hyperbola centered at the origin, the length of the transverse axis is $2a$.

So, $2a = 4$.

Dividing by 2, we get $a = 2$.

Squaring both sides, $a^2 = 2^2 = 4$.

For a hyperbola, the relationship between $a$, $b$, and $c$ is $c^2 = a^2 + b^2$.

We can use this relationship to find $b^2$:

$5 = 4 + b^2$

$b^2 = 5 - 4$

$b^2 = 1$

Now, substitute the values of $a^2 = 4$ and $b^2 = 1$ into the standard equation of the horizontal hyperbola:

$\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$

$\frac{x^2}{4} - \frac{y^2}{1} = 1$

$\frac{x^2}{4} - y^2 = 1$

The equation of the hyperbola is $\mathbf{\frac{x^2}{4} - y^2 = 1}$.

Given:

The equation of the parabola is $y^2 = 8x$.

To Find:

The latus rectum and eccentricity of the parabola.

Solution:

The given equation of the parabola is $y^2 = 8x$.

This equation is in the standard form $y^2 = 4ax$, which represents a parabola with vertex at the origin and opening towards the positive x-axis.

Comparing $y^2 = 8x$ with $y^2 = 4ax$, we have:

$4a = 8$

$a = \frac{8}{4}$

$a = 2$

The length of the latus rectum of a parabola of the form $y^2 = 4ax$ is given by $4a$.

Latus Rectum $= 4a$

Substitute the value of $a=2$:

Latus Rectum $= 4(2)$

Latus Rectum $= 8$

The eccentricity of any parabola is always equal to 1.

Eccentricity $e = 1$

Answer:

The latus rectum of the parabola $y^2 = 8x$ is $\mathbf{8}$.

The eccentricity of the parabola $y^2 = 8x$ is $\mathbf{1}$.

Given:

A point on the circle is $(2, 3)$.

The centre of the circle is $(-1, 4)$.

To Find:

The equation of the circle.

Solution:

The standard equation of a circle with centre $(h, k)$ and radius $r$ is given by:

$(x - h)^2 + (y - k)^2 = r^2$

The given centre of the circle is $(h, k) = (-1, 4)$.

To find the equation, we first need to find the radius $r$. The radius is the distance between the centre and any point on the circle.

Using the distance formula between the centre $(-1, 4)$ and the point on the circle $(2, 3)$:

$r = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$

Let $(x_1, y_1) = (-1, 4)$ and $(x_2, y_2) = (2, 3)$.

$r = \sqrt{(2 - (-1))^2 + (3 - 4)^2}$

$r = \sqrt{(2 + 1)^2 + (-1)^2}$

$r = \sqrt{3^2 + 1}$

$r = \sqrt{9 + 1}$

$r = \sqrt{10}$

Now, we find $r^2$:

$r^2 = (\sqrt{10})^2 = 10$

Substitute the coordinates of the centre $(h, k) = (-1, 4)$ and the value of $r^2 = 10$ into the standard equation of the circle:

$(x - (-1))^2 + (y - 4)^2 = 10$

$(x + 1)^2 + (y - 4)^2 = 10$

This is the equation of the circle.

We can also expand this equation to the general form:

$(x^2 + 2x + 1) + (y^2 - 8y + 16) = 10$

$x^2 + y^2 + 2x - 8y + 1 + 16 = 10$

$x^2 + y^2 + 2x - 8y + 17 = 10$

$x^2 + y^2 + 2x - 8y + 17 - 10 = 0$

$x^2 + y^2 + 2x - 8y + 7 = 0$

Answer:

The equation of the circle is $\mathbf{(x + 1)^2 + (y - 4)^2 = 10}$ or in general form $\mathbf{x^2 + y^2 + 2x - 8y + 7 = 0}$.

Given:

The equation of the ellipse is $\frac{x^2}{25} + \frac{y^2}{9} = 1$.

To Find:

The vertices and foci of the ellipse.

Solution:

The given equation of the ellipse is $\frac{x^2}{25} + \frac{y^2}{9} = 1$.

This equation is in the standard form $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ or $\frac{x^2}{b^2} + \frac{y^2}{a^2} = 1$ for an ellipse centered at the origin $(0, 0)$.

Comparing the denominators, we have $25$ and $9$. Since $25 > 9$, the larger denominator is under the $x^2$ term.

This indicates that the major axis is along the x-axis, and the ellipse is horizontal.

For a horizontal ellipse centered at the origin, the standard form is $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$, where $a^2$ is the larger denominator.

From the given equation, we have:

$a^2 = 25$

$b^2 = 9$

Taking the square root of $a^2$ and $b^2$ (considering only positive values for $a$ and $b$ as lengths):

$a = \sqrt{25} = 5$

$b = \sqrt{9} = 3$

For a horizontal ellipse centered at the origin, the vertices are located at $(\pm a, 0)$.

Vertices: $(\pm 5, 0)$

To find the foci, we need to find the value of $c$ using the relationship $c^2 = a^2 - b^2$ for an ellipse.

$c^2 = 25 - 9$

$c^2 = 16$

$c = \sqrt{16} = 4$

For a horizontal ellipse centered at the origin, the foci are located at $(\pm c, 0)$.

Foci: $(\pm 4, 0)$

Answer:

The vertices of the ellipse are $\mathbf{(\pm 5, 0)}$.

The foci of the ellipse are $\mathbf{(\pm 4, 0)}$.

Given:

Vertices of the hyperbola are $(\pm 2, 0)$.

Foci of the hyperbola are $(\pm 3, 0)$.

To Find:

The equation of the hyperbola.

Solution:

The vertices are given as $(\pm 2, 0)$ and the foci are given as $(\pm 3, 0)$.

Since the vertices and foci are on the x-axis and are symmetric with respect to the origin, the center of the hyperbola is at the origin $(0, 0)$.

The standard equation of a hyperbola centered at the origin is $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$ for a horizontal transverse axis or $\frac{y^2}{a^2} - \frac{x^2}{b^2} = 1$ for a vertical transverse axis.

The distance from the center to a vertex is denoted by $a$. From the vertices $(\pm 2, 0)$, we have $a = 2$.

$a^2 = 2^2 = 4$.

The distance from the center to a focus is denoted by $c$. From the foci $(\pm 3, 0)$, we have $c = 3$.

$c^2 = 3^2 = 9$.

Since the vertices $(\pm a, 0)$ and foci $(\pm c, 0)$ are on the x-axis, the transverse axis is horizontal.

The standard equation for a horizontal hyperbola centered at the origin is:

$\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$

For a hyperbola, the relationship between $a$, $b$, and $c$ is $c^2 = a^2 + b^2$.

We can use this relationship to find $b^2$:

$b^2 = c^2 - a^2$

Substitute the values $c^2 = 9$ and $a^2 = 4$:

$b^2 = 9 - 4$

$b^2 = 5$

Now, substitute the values of $a^2 = 4$ and $b^2 = 5$ into the standard equation of the horizontal hyperbola:

$\frac{x^2}{4} - \frac{y^2}{5} = 1$

The equation of the hyperbola is $\mathbf{\frac{x^2}{4} - \frac{y^2}{5} = 1}$.

Given:

The circle passes through the points $(0, 0)$, $(a, 0)$, and $(0, b)$.

To Find:

The equation of the circle.

Solution:

The general equation of a circle is given by:

$x^2 + y^2 + 2gx + 2fy + c = 0$

where $(-g, -f)$ is the centre and $\sqrt{g^2 + f^2 - c}$ is the radius.

Since the circle passes through the point $(0, 0)$, we can substitute $x = 0$ and $y = 0$ into the general equation:

$0^2 + 0^2 + 2g(0) + 2f(0) + c = 0$

$c = 0$

Now the equation of the circle becomes:

$x^2 + y^2 + 2gx + 2fy = 0$

Since the circle passes through the point $(a, 0)$, substitute $x = a$ and $y = 0$ into the modified equation:

$a^2 + 0^2 + 2g(a) + 2f(0) = 0$

$a^2 + 2ga = 0$

$a(a + 2g) = 0$

Since $(a, 0)$ is a point on the circle and distinct from $(0, 0)$, we assume $a \neq 0$. Therefore,

$a + 2g = 0$

$2g = -a$

$g = -\frac{a}{2}$

Since the circle passes through the point $(0, b)$, substitute $x = 0$ and $y = b$ into the equation $x^2 + y^2 + 2gx + 2fy = 0$:

$0^2 + b^2 + 2g(0) + 2f(b) = 0$

$b^2 + 2fb = 0$

$b(b + 2f) = 0$

Since $(0, b)$ is a point on the circle and distinct from $(0, 0)$, we assume $b \neq 0$. Therefore,

$b + 2f = 0$

$2f = -b$

$f = -\frac{b}{2}$

Now, substitute the values of $g = -\frac{a}{2}$, $f = -\frac{b}{2}$, and $c = 0$ back into the general equation of the circle $x^2 + y^2 + 2gx + 2fy + c = 0$:

$x^2 + y^2 + 2\left(-\frac{a}{2}\right)x + 2\left(-\frac{b}{2}\right)y + 0 = 0$

$x^2 + y^2 - ax - by = 0$

The equation of the circle passing through the points $(0, 0)$, $(a, 0)$, and $(0, b)$ is $\mathbf{x^2 + y^2 - ax - by = 0}$.

Given:

Focus F = $(1, -1)$

Directrix equation = $x + y + 7 = 0$

To Find:

The equation of the parabola.

Solution:

A parabola is the locus of a point P(x, y) such that its distance from the focus is equal to its distance from the directrix.

Let P be any point $(x, y)$ on the parabola.

The distance of the point P from the focus F(1, -1) is given by the distance formula:

$PF = \sqrt{(x - 1)^2 + (y - (-1))^2} = \sqrt{(x - 1)^2 + (y + 1)^2}$

The distance of the point P(x, y) from the directrix $x + y + 7 = 0$ is given by the formula for the distance from a point to a line $|Ax_0 + By_0 + C| / \sqrt{A^2 + B^2}$.

Here, $(x_0, y_0) = (x, y)$, $A = 1$, $B = 1$, and $C = 7$.

Distance from P to directrix $= \frac{|x + y + 7|}{\sqrt{1^2 + 1^2}} = \frac{|x + y + 7|}{\sqrt{2}}$

According to the definition of a parabola, $PF =$ Distance from P to directrix.

$\sqrt{(x - 1)^2 + (y + 1)^2} = \frac{|x + y + 7|}{\sqrt{2}}$

Square both sides of the equation to eliminate the square root and the absolute value:

$(x - 1)^2 + (y + 1)^2 = \left(\frac{x + y + 7}{\sqrt{2}}\right)^2$

$(x^2 - 2x + 1) + (y^2 + 2y + 1) = \frac{(x + y + 7)^2}{2}$

$x^2 + y^2 - 2x + 2y + 2 = \frac{(x + y + 7)^2}{2}$

Expand the term $(x + y + 7)^2$ using the formula $(a + b + c)^2 = a^2 + b^2 + c^2 + 2ab + 2ac + 2bc$:

$(x + y + 7)^2 = x^2 + y^2 + 7^2 + 2(x)(y) + 2(x)(7) + 2(y)(7)$

$(x + y + 7)^2 = x^2 + y^2 + 49 + 2xy + 14x + 14y$

Substitute this back into the equation:

$x^2 + y^2 - 2x + 2y + 2 = \frac{x^2 + y^2 + 2xy + 14x + 14y + 49}{2}$

Multiply both sides by 2:

$2(x^2 + y^2 - 2x + 2y + 2) = x^2 + y^2 + 2xy + 14x + 14y + 49$

$2x^2 + 2y^2 - 4x + 4y + 4 = x^2 + y^2 + 2xy + 14x + 14y + 49$

Move all terms to the left side and combine like terms:

$2x^2 - x^2 + 2y^2 - y^2 - 2xy - 4x - 14x + 4y - 14y + 4 - 49 = 0$

$x^2 + y^2 - 2xy - 18x - 10y - 45 = 0$

The equation of the parabola is $\mathbf{x^2 - 2xy + y^2 - 18x - 10y - 45 = 0}$.

Given:

The equation of the ellipse is $\frac{x^2}{16} + \frac{y^2}{25} = 1$.

To Find:

The lengths of the major and minor axes, the coordinates of the foci, the vertices, and the eccentricity of the ellipse.

Solution:

The given equation of the ellipse is $\frac{x^2}{16} + \frac{y^2}{25} = 1$.

This equation is in the standard form $\frac{x^2}{b^2} + \frac{y^2}{a^2} = 1$ or $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ for an ellipse centered at the origin $(0, 0)$.

Comparing the denominators, we have $16$ and $25$. Since $25 > 16$, the larger denominator is under the $y^2$ term.

This indicates that the major axis is along the y-axis, and the ellipse is vertical.

For a vertical ellipse centered at the origin, the standard form is $\frac{x^2}{b^2} + \frac{y^2}{a^2} = 1$, where $a^2$ is the larger denominator.

From the given equation, we have:

$a^2 = 25$

$b^2 = 16$

Taking the square root of $a^2$ and $b^2$ (considering only positive values for $a$ and $b$ as lengths):

$a = \sqrt{25} = 5$

$b = \sqrt{16} = 4$

Length of the major axis:

The length of the major axis is $2a$.

Length of major axis $= 2 \times 5 = \mathbf{10}$.

Length of the minor axis:

The length of the minor axis is $2b$.

Length of minor axis $= 2 \times 4 = \mathbf{8}$.

Coordinates of the vertices:

For a vertical ellipse centered at the origin, the vertices are located at $(0, \pm a)$.

Vertices: $(0, \pm 5)$. The vertices are $\mathbf{(0, 5)}$ and $\mathbf{(0, -5)}$.

Coordinates of the foci:

To find the foci, we need to find the value of $c$ using the relationship $c^2 = a^2 - b^2$ for an ellipse.

$c^2 = 25 - 16$

$c^2 = 9$

$c = \sqrt{9} = 3$

For a vertical ellipse centered at the origin, the foci are located at $(0, \pm c)$.

Foci: $(0, \pm 3)$. The foci are $\mathbf{(0, 3)}$ and $\mathbf{(0, -3)}$.

Eccentricity:

The eccentricity $e$ of an ellipse is given by $e = \frac{c}{a}$.

$e = \frac{3}{5}$

Eccentricity $e = \mathbf{\frac{3}{5}}$.

Summary of results:

Length of major axis: $\mathbf{10}$

Length of minor axis: $\mathbf{8}$

Vertices: $\mathbf{(0, \pm 5)}$

Foci: $\mathbf{(0, \pm 3)}$

Eccentricity: $\mathbf{\frac{3}{5}}$

Given:

The equation of the hyperbola is $9y^2 - 4x^2 = 36$.

To Find:

The eccentricity of the hyperbola.

Solution:

The given equation of the hyperbola is $9y^2 - 4x^2 = 36$.

To find the eccentricity, we first need to write the equation in the standard form.

Divide the entire equation by 36:

$\frac{9y^2}{36} - \frac{4x^2}{36} = \frac{36}{36}$

$\frac{y^2}{4} - \frac{x^2}{9} = 1$

This equation is in the standard form of a hyperbola with a vertical transverse axis, which is $\frac{y^2}{a^2} - \frac{x^2}{b^2} = 1$.

Comparing the given equation with the standard form, we have:

$a^2 = 4 \implies a = \sqrt{4} = 2$

$b^2 = 9 \implies b = \sqrt{9} = 3$

For a hyperbola, the relationship between $a$, $b$, and $c$ (where $c$ is the distance from the center to the focus) is given by $c^2 = a^2 + b^2$.

$c^2 = 4 + 9$

$c^2 = 13$

$c = \sqrt{13}$

The eccentricity $e$ of a hyperbola is defined as the ratio $e = \frac{c}{a}$.

Substitute the values of $c = \sqrt{13}$ and $a = 2$:

$e = \frac{\sqrt{13}}{2}$

Answer:

The eccentricity of the hyperbola $9y^2 - 4x^2 = 36$ is $\mathbf{\frac{\sqrt{13}}{2}}$.

Given:

The circle passes through points P(1, -2) and Q(4, -3).

The centre of the circle lies on the line $3x + 4y = 7$.

To Find:

The equation of the circle.

Solution:

Let the centre of the circle be C$(h, k)$ and the radius be $r$.

The standard equation of a circle with centre $(h, k)$ and radius $r$ is $(x - h)^2 + (y - k)^2 = r^2$.

Since the points P(1, -2) and Q(4, -3) lie on the circle, their distances from the centre C$(h, k)$ must be equal to the radius $r$.

Distance CP = $r$: $(1 - h)^2 + (-2 - k)^2 = r^2$

Distance CQ = $r$: $(4 - h)^2 + (-3 - k)^2 = r^2$

Equating the two expressions for $r^2$:

$(1 - h)^2 + (-2 - k)^2 = (4 - h)^2 + (-3 - k)^2$

Expand both sides:

$(1 - 2h + h^2) + (4 + 4k + k^2) = (16 - 8h + h^2) + (9 + 6k + k^2)$

$1 - 2h + h^2 + 4 + 4k + k^2 = 16 - 8h + h^2 + 9 + 6k + k^2$

Combine like terms:

$h^2 + k^2 - 2h + 4k + 5 = h^2 + k^2 - 8h + 6k + 25$

Subtract $h^2 + k^2$ from both sides and rearrange the terms:

$-2h + 8h + 4k - 6k + 5 - 25 = 0$}

$6h - 2k - 20 = 0$

Divide by 2:

$3h - k - 10 = 0$

So, we get the equation: $3h - k = 10$}

The centre C$(h, k)$ lies on the line $3x + 4y = 7$. Substituting the coordinates of the centre into the line equation gives:

$3h + 4k = 7$}

Now we have a system of two linear equations for $h$ and $k$:

1) $3h - k = 10$

2) $3h + 4k = 7$

From equation (1), we can express $k$ in terms of $h$: $k = 3h - 10$.

Substitute this expression for $k$ into equation (2):

$3h + 4(3h - 10) = 7$

$3h + 12h - 40 = 7$

$15h = 47$

$h = \frac{47}{15}$

Substitute the value of $h$ back into the expression for $k$:

$k = 3\left(\frac{47}{15}\right) - 10$

$k = \frac{47}{5} - 10$

$k = \frac{47 - 50}{5}$

$k = -\frac{3}{5}$

The centre of the circle is $\mathbf{\left(\frac{47}{15}, -\frac{3}{5}\right)}$.

Now we calculate the radius squared, $r^2$, by substituting the coordinates of the centre $(h, k)$ and one of the points on the circle, say (1, -2), into the distance formula (or the squared distance formula):

$r^2 = (1 - h)^2 + (-2 - k)^2$

$r^2 = \left(1 - \frac{47}{15}\right)^2 + \left(-2 - \left(-\frac{3}{5}\right)\right)^2$

$r^2 = \left(\frac{15 - 47}{15}\right)^2 + \left(-2 + \frac{3}{5}\right)^2$

$r^2 = \left(\frac{-32}{15}\right)^2 + \left(\frac{-10 + 3}{5}\right)^2$

$r^2 = \left(\frac{-32}{15}\right)^2 + \left(\frac{-7}{5}\right)^2$

$r^2 = \frac{(-32)^2}{15^2} + \frac{(-7)^2}{5^2}$

$r^2 = \frac{1024}{225} + \frac{49}{25}$

To add the fractions, find a common denominator (225):

$r^2 = \frac{1024}{225} + \frac{49 \times 9}{25 \times 9}$

$r^2 = \frac{1024}{225} + \frac{441}{225}$

$r^2 = \frac{1024 + 441}{225}$

$r^2 = \frac{1465}{225}$

Now, write the equation of the circle using the standard form $(x - h)^2 + (y - k)^2 = r^2$ with $h = \frac{47}{15}$, $k = -\frac{3}{5}$, and $r^2 = \frac{1465}{225}$:

$\left(x - \frac{47}{15}\right)^2 + \left(y - \left(-\frac{3}{5}\right)\right)^2 = \frac{1465}{225}$

$\left(x - \frac{47}{15}\right)^2 + \left(y + \frac{3}{5}\right)^2 = \frac{1465}{225}$

Alternatively, we can express the equation in the general form $x^2 + y^2 + 2gx + 2fy + c = 0$, where $g = -h = -\frac{47}{15}$ and $f = -k = \frac{3}{5}$.

The constant term $c$ is given by $c = h^2 + k^2 - r^2$ or $c = g^2 + f^2 - r^2$.

$c = \left(\frac{47}{15}\right)^2 + \left(-\frac{3}{5}\right)^2 - \frac{1465}{225}$

$c = \frac{2209}{225} + \frac{9}{25} - \frac{1465}{225}$

$c = \frac{2209}{225} + \frac{81}{225} - \frac{1465}{225}$

$c = \frac{2209 + 81 - 1465}{225} = \frac{2290 - 1465}{225} = \frac{825}{225} = \frac{11}{3}$

Substituting the values of $g$, $f$, and $c$ into the general equation:

$x^2 + y^2 + 2\left(-\frac{47}{15}\right)x + 2\left(\frac{3}{5}\right)y + \frac{11}{3} = 0$

$x^2 + y^2 - \frac{94}{15}x + \frac{6}{5}y + \frac{11}{3} = 0$}

Multiply the equation by the LCM of the denominators (15) to get integer coefficients:

$15(x^2 + y^2) - 15\left(\frac{94}{15}\right)x + 15\left(\frac{6}{5}\right)y + 15\left(\frac{11}{3}\right) = 15(0)$

$15x^2 + 15y^2 - 94x + 18y + 55 = 0$

Answer:

The equation of the circle is $\mathbf{\left(x - \frac{47}{15}\right)^2 + \left(y + \frac{3}{5}\right)^2 = \frac{1465}{225}}$ or $\mathbf{15x^2 + 15y^2 - 94x + 18y + 55 = 0}$.

Given:

Length of the latus rectum = 12

Vertex of the parabola is at the origin $(0, 0)$.

The axis of the parabola is along the positive x-axis.

To Find:

The equation of the parabola.

Solution:

Since the vertex is at the origin $(0, 0)$ and the axis is along the positive x-axis, the standard equation of the parabola is of the form $y^2 = 4ax$, where $a > 0$.

For a parabola of the form $y^2 = 4ax$, the length of the latus rectum is given by $4a$.

We are given that the length of the latus rectum is 12.

Divide both sides by 4 to find the value of $a$:

$a = \frac{12}{4}$

$a = 3$

Substitute the value of $a = 3$ into the standard equation of the parabola $y^2 = 4ax$:

$y^2 = 4(3)x$

$y^2 = 12x$

The equation of the parabola is $\mathbf{y^2 = 12x}$.

Given:

The major axis of the ellipse is along the x-axis.

The ellipse passes through the points (4, 3) and (6, 2).

To Find:

The equation of the ellipse.

Solution:

Since the major axis is along the x-axis, the standard equation of the ellipse centered at the origin is of the form:

$\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$

where $a$ is the length of the semi-major axis and $b$ is the length of the semi-minor axis ($a > b$).

Since the ellipse passes through the point (4, 3), we substitute $x = 4$ and $y = 3$ into the equation:

$\frac{4^2}{a^2} + \frac{3^2}{b^2} = 1$

$\frac{16}{a^2} + \frac{9}{b^2} = 1$

Since the ellipse passes through the point (6, 2), we substitute $x = 6$ and $y = 2$ into the equation:

$\frac{6^2}{a^2} + \frac{2^2}{b^2} = 1$

$\frac{36}{a^2} + \frac{4}{b^2} = 1$

We now have a system of two linear equations in terms of $\frac{1}{a^2}$ and $\frac{1}{b^2}$. Let $X = \frac{1}{a^2}$ and $Y = \frac{1}{b^2}$.

1) $16X + 9Y = 1$}

2) $36X + 4Y = 1$}

We can solve this system using elimination or substitution. Let's use elimination.

Multiply equation (1) by 4 and equation (2) by 9 to make the coefficients of Y equal:

$4 \times (16X + 9Y) = 4 \times 1 \implies 64X + 36Y = 4$

$9 \times (36X + 4Y) = 9 \times 1 \implies 324X + 36Y = 9$

Subtract the first new equation from the second new equation:

$(324X + 36Y) - (64X + 36Y) = 9 - 4$

$324X - 64X + 36Y - 36Y = 5$

$260X = 5$

$X = \frac{5}{260} = \frac{1}{52}$

So, $\frac{1}{a^2} = \frac{1}{52}$, which means $a^2 = 52$.

Now substitute the value of $X = \frac{1}{52}$ into equation (1) to find Y:

$16\left(\frac{1}{52}\right) + 9Y = 1$

$\frac{16}{52} + 9Y = 1$

Simplify the fraction $\frac{16}{52}$: $\frac{\cancel{16}^{4}}{\cancel{52}_{13}} = \frac{4}{13}$

$\frac{4}{13} + 9Y = 1$

$9Y = 1 - \frac{4}{13}$

$9Y = \frac{13 - 4}{13} = \frac{9}{13}$

$Y = \frac{9}{13} \times \frac{1}{9} = \frac{1}{13}$

So, $\frac{1}{b^2} = \frac{1}{13}$, which means $b^2 = 13$.

Substitute the values of $a^2 = 52$ and $b^2 = 13$ into the standard equation of the ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$:

$\frac{x^2}{52} + \frac{y^2}{13} = 1$

We should verify that $a^2 > b^2$, which is $52 > 13$. This is true, confirming the major axis is along the x-axis as given.

Answer:

The equation of the ellipse is $\mathbf{\frac{x^2}{52} + \frac{y^2}{13} = 1}$.

Given:

The equation of the hyperbola is $\frac{x^2}{16} - \frac{y^2}{9} = 1$.

To Find:

The foci, vertices, and eccentricity of the hyperbola.

Solution:

The given equation of the hyperbola is $\frac{x^2}{16} - \frac{y^2}{9} = 1$.

This equation is in the standard form of a hyperbola centered at the origin with a horizontal transverse axis, which is $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$.

Comparing the given equation with the standard form, we have:

$a^2 = 16 \implies a = \sqrt{16} = 4$

$b^2 = 9 \implies b = \sqrt{9} = 3$

Vertices:

For a horizontal hyperbola centered at the origin, the vertices are located at $(\pm a, 0)$.

Vertices: $(\pm 4, 0)$. The vertices are $\mathbf{(4, 0)}$ and $\mathbf{(-4, 0)}$.

Foci:

To find the foci, we need to find the value of $c$ using the relationship $c^2 = a^2 + b^2$ for a hyperbola.

$c^2 = 16 + 9$

$c^2 = 25$

$c = \sqrt{25} = 5$

For a horizontal hyperbola centered at the origin, the foci are located at $(\pm c, 0)$.

Foci: $(\pm 5, 0)$. The foci are $\mathbf{(5, 0)}$ and $\mathbf{(-5, 0)}$.

Eccentricity:

The eccentricity $e$ of a hyperbola is defined as the ratio $e = \frac{c}{a}$.

Substitute the values of $c = 5$ and $a = 4$:

$e = \frac{5}{4}$

Eccentricity $e = \mathbf{\frac{5}{4}}$.

Summary of results:

Vertices: $\mathbf{(\pm 4, 0)}$

Foci: $\mathbf{(\pm 5, 0)}$

Eccentricity: $\mathbf{\frac{5}{4}}$

Given:

The required circle is concentric with the circle $x^2 + y^2 - 6x + 12y + 15 = 0$.

The area of the required circle is 154 square units.

Use $\pi = \frac{22}{7}$.

To Find:

The equation of the required circle.

Solution:

Two circles are concentric if they have the same centre.

The general equation of a circle is $x^2 + y^2 + 2gx + 2fy + c = 0$. The centre of this circle is $(-g, -f)$.

The given circle is $x^2 + y^2 - 6x + 12y + 15 = 0$.

Comparing this with the general equation, we have $2g = -6$ and $2f = 12$.

$g = -3$

$f = 6$

The centre of the given circle is $(-g, -f) = (-(-3), -6) = (3, -6)$.

Since the required circle is concentric with this circle, its centre is also $(3, -6)$.

Let the centre of the required circle be $(h, k) = (3, -6)$.

Let the radius of the required circle be $R$.

The area of the required circle is given as 154 square units.

The formula for the area of a circle is Area $= \pi R^2$.

Substitute the given value of $\pi = \frac{22}{7}$:

$R^2 = 154 \times \frac{7}{22}$

$R^2 = \cancel{154}^{7} \times \frac{7}{\cancel{22}_{1}}$

$R^2 = 7 \times 7$

$R^2 = 49$

The equation of a circle with centre $(h, k)$ and radius $R$ is $(x - h)^2 + (y - k)^2 = R^2$.

Substitute the centre $(h, k) = (3, -6)$ and $R^2 = 49$ into the equation:

$(x - 3)^2 + (y - (-6))^2 = 49$

$(x - 3)^2 + (y + 6)^2 = 49$

This is the equation of the required circle in standard form.

We can also expand it into the general form:

$(x^2 - 6x + 9) + (y^2 + 12y + 36) = 49$

$x^2 + y^2 - 6x + 12y + 9 + 36 = 49$

$x^2 + y^2 - 6x + 12y + 45 = 49$

$x^2 + y^2 - 6x + 12y + 45 - 49 = 0$

$x^2 + y^2 - 6x + 12y - 4 = 0$

Answer:

The equation of the circle is $\mathbf{(x - 3)^2 + (y + 6)^2 = 49}$ or $\mathbf{x^2 + y^2 - 6x + 12y - 4 = 0}$.

Given:

The axis of the parabola is the y-axis.

The vertex of the parabola is at the origin $(0, 0)$.

The parabola passes through the point (5, 2).

To Find:

The equation of the parabola.

Solution:

Since the vertex is at the origin $(0, 0)$ and the axis is along the y-axis, the standard equation of the parabola is of the form $x^2 = 4ay$ (opening upwards) or $x^2 = -4ay$ (opening downwards).

The parabola passes through the point (5, 2).

Substitute the coordinates of the point (5, 2) into the possible standard equations to determine the correct form and the value of $a$.

Consider the equation $x^2 = 4ay$. Substitute $x=5$ and $y=2$:

$5^2 = 4a(2)$

$25 = 8a$

$a = \frac{25}{8}$

Since the point (5, 2) has positive x and y coordinates, and the vertex is at the origin, the parabola must open upwards, meaning $a$ must be positive. The value $a = \frac{25}{8}$ is positive.

So, the equation is of the form $x^2 = 4ay$ with $a = \frac{25}{8}$.

Substitute the value of $a$ back into the equation $x^2 = 4ay$:

$x^2 = 4\left(\frac{25}{8}\right)y$

$x^2 = \cancel{4}^{1} \times \frac{25}{\cancel{8}_{2}}y$

$x^2 = \frac{25}{2}y$

Multiply both sides by 2:

$2x^2 = 25y$

Or, $2x^2 - 25y = 0$

If we considered the equation $x^2 = -4ay$ and substituted $(5, 2)$: $5^2 = -4a(2) \implies 25 = -8a \implies a = -\frac{25}{8}$. A parabola opening downwards ($x^2 = -4ay$) would have a positive value for $a$ in the standard form $x^2 = -4ay$. The calculated value of $a = -\frac{25}{8}$ would mean $-4a = -4(-\frac{25}{8}) = \frac{25}{2}$, leading to $x^2 = \frac{25}{2}y$, which is the same equation. The interpretation of $a$ depends on the initial standard form chosen. With vertex at origin and axis along y-axis, the forms are $x^2=4ay$ and $x^2=-4ay$. Since $(5,2)$ is in the first quadrant, the parabola must open upwards ($x^2=4ay$) as it passes through $(5,2)$ and its vertex is at origin. Thus $a > 0$.

Answer:

The equation of the parabola is $\mathbf{x^2 = \frac{25}{2}y}$ or $\mathbf{2x^2 = 25y}$.

Given:

Foci of the ellipse are $(\pm 3, 0)$.

Length of the latus rectum is 8.

To Find:

The equation of the ellipse.

Solution:

The foci are given as $(\pm 3, 0)$. Since the foci are on the x-axis and are symmetric with respect to the origin, the center of the ellipse is at the origin $(0, 0)$ and the major axis is along the x-axis.

The standard equation of a horizontal ellipse centered at the origin is:

$\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$

where $a$ is the length of the semi-major axis and $b$ is the length of the semi-minor axis, with $a > b$.

From the foci $(\pm 3, 0)$, the distance from the center to a focus is $c = 3$.

So, $c^2 = 3^2 = 9$.

For an ellipse with the major axis along the x-axis, the length of the latus rectum is given by $\frac{2b^2}{a}$.

We are given that the length of the latus rectum is 8.

Multiply both sides by $a$ and divide by 2:

$2b^2 = 8a$

$b^2 = 4a$

For an ellipse, the relationship between $a$, $b$, and $c$ is $c^2 = a^2 - b^2$ (since $a > b$).

Substitute the values $c^2 = 9$ and $b^2 = 4a$ into this relationship:

$9 = a^2 - 4a$

Rearrange the terms to form a quadratic equation in $a$:

$a^2 - 4a - 9 = 0$

We can solve this quadratic equation for $a$ using the quadratic formula $a = \frac{-B \pm \sqrt{B^2 - 4AC}}{2A}$, where $A=1$, $B=-4$, and $C=-9$.

$a = \frac{-(-4) \pm \sqrt{(-4)^2 - 4(1)(-9)}}{2(1)}$

$a = \frac{4 \pm \sqrt{16 + 36}}{2}$

$a = \frac{4 \pm \sqrt{52}}{2}$

Simplify the square root: $\sqrt{52} = \sqrt{4 \times 13} = \sqrt{4} \times \sqrt{13} = 2\sqrt{13}$.

$a = \frac{4 \pm 2\sqrt{13}}{2}$

$a = 2 \pm \sqrt{13}$

Since $a$ represents the length of the semi-major axis, it must be a positive value. Also, for the foci to be at $(\pm 3, 0)$ on the x-axis, the vertices must be at $(\pm a, 0)$ with $|a| > |c|$, i.e., $a > 3$.

Consider the two possible values for $a$:

$a_1 = 2 + \sqrt{13}$

$a_2 = 2 - \sqrt{13}$

Since $\sqrt{13} \approx 3.6$, $a_1 = 2 + \sqrt{13} \approx 2 + 3.6 = 5.6$, which is positive and greater than 3. $a_2 = 2 - \sqrt{13} \approx 2 - 3.6 = -1.6$, which is negative and not a valid length.

Therefore, we take the positive value $a = 2 + \sqrt{13}$.

Now, we find $a^2$:

$a^2 = (2 + \sqrt{13})^2 = 2^2 + 2(2)(\sqrt{13}) + (\sqrt{13})^2$

$a^2 = 4 + 4\sqrt{13} + 13$

$a^2 = 17 + 4\sqrt{13}$

Next, find $b^2$ using the relation $b^2 = 4a$:

$b^2 = 4(2 + \sqrt{13})$

$b^2 = 8 + 4\sqrt{13}$

Substitute the values of $a^2$ and $b^2$ into the standard equation of the ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$:

$\frac{x^2}{17 + 4\sqrt{13}} + \frac{y^2}{8 + 4\sqrt{13}} = 1$

Answer:

The equation of the ellipse is $\mathbf{\frac{x^2}{17 + 4\sqrt{13}} + \frac{y^2}{8 + 4\sqrt{13}} = 1}$.

Given:

Vertices of the hyperbola are $(\pm 7, 0)$.

Eccentricity $e = \frac{4}{3}$.

To Find:

The equation of the hyperbola.

Solution:

The vertices are given as $(\pm 7, 0)$. Since the vertices are on the x-axis and are symmetric with respect to the origin, the center of the hyperbola is at the origin $(0, 0)$ and the transverse axis is along the x-axis.

The standard equation of a horizontal hyperbola centered at the origin is:

$\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$

The distance from the center to a vertex is denoted by $a$. From the vertices $(\pm 7, 0)$, we have $a = 7$.

$a^2 = 7^2 = 49$.

The eccentricity $e$ of a hyperbola is defined as $e = \frac{c}{a}$, where $c$ is the distance from the center to the focus.

We are given $e = \frac{4}{3}$ and we found $a = 7$.

Solve for $c$:

$c = \frac{4}{3} \times 7 = \frac{28}{3}$.

For a hyperbola, the relationship between $a$, $b$, and $c$ is $c^2 = a^2 + b^2$.

We can use this relationship to find $b^2$:

$b^2 = c^2 - a^2$

Substitute the values $c = \frac{28}{3}$ and $a = 7$:

$b^2 = \left(\frac{28}{3}\right)^2 - 7^2$

$b^2 = \frac{28^2}{3^2} - 49$

$b^2 = \frac{784}{9} - 49$

Convert 49 to a fraction with denominator 9: $49 = \frac{49 \times 9}{9} = \frac{441}{9}$.

$b^2 = \frac{784}{9} - \frac{441}{9}$

$b^2 = \frac{784 - 441}{9} = \frac{343}{9}$

Now, substitute the values of $a^2 = 49$ and $b^2 = \frac{343}{9}$ into the standard equation of the horizontal hyperbola $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$:

$\frac{x^2}{49} - \frac{y^2}{\frac{343}{9}} = 1$}

Simplify the term with $b^2$ in the denominator:

$\frac{x^2}{49} - \frac{9y^2}{343} = 1$}

Answer:

The equation of the hyperbola is $\mathbf{\frac{x^2}{49} - \frac{9y^2}{343} = 1}$.

Given:

The diameter of the circle is the line segment joining the points P(-4, 3) and Q(12, -1).

To Find:

The equation of the circle.

Solution:

The center of the circle is the midpoint of its diameter. Let the center be C$(h, k)$.

Using the midpoint formula, the coordinates of the center are:

$h = \frac{x_1 + x_2}{2} = \frac{-4 + 12}{2} = \frac{8}{2} = 4$

$k = \frac{y_1 + y_2}{2} = \frac{3 + (-1)}{2} = \frac{2}{2} = 1$

The center of the circle is $\mathbf{(4, 1)}$.

The radius of the circle is the distance from the center to either endpoint of the diameter. Let's use the point (12, -1).

Let the radius be $r$. Using the distance formula, $r$ is the distance between C(4, 1) and Q(12, -1).

$r = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$

$r = \sqrt{(12 - 4)^2 + (-1 - 1)^2}$

$r = \sqrt{8^2 + (-2)^2}$

$r = \sqrt{64 + 4}$

$r = \sqrt{68}$

The standard equation of a circle with center $(h, k)$ and radius $r$ is $(x - h)^2 + (y - k)^2 = r^2$.

Substitute the center $(h, k) = (4, 1)$ and $r^2 = (\sqrt{68})^2 = 68$ into the standard equation:

$(x - 4)^2 + (y - 1)^2 = 68$

This is the equation of the circle. We can also expand it into the general form:

$(x^2 - 8x + 16) + (y^2 - 2y + 1) = 68$

$x^2 + y^2 - 8x - 2y + 16 + 1 = 68$

$x^2 + y^2 - 8x - 2y + 17 = 68$

$x^2 + y^2 - 8x - 2y + 17 - 68 = 0$

$x^2 + y^2 - 8x - 2y - 51 = 0$

Answer:

The equation of the circle is $\mathbf{(x - 4)^2 + (y - 1)^2 = 68}$ or $\mathbf{x^2 + y^2 - 8x - 2y - 51 = 0}$.

Given:

The circle passes through the points A(2, 0), B(0, 2), and O(0, 0).

To Find:

The equation of the circle.

Solution:

The general equation of a circle is given by:

$x^2 + y^2 + 2gx + 2fy + c = 0$

Since the circle passes through the point O(0, 0), we substitute $x = 0$ and $y = 0$ into the general equation:

$0^2 + 0^2 + 2g(0) + 2f(0) + c = 0$

Now the equation of the circle becomes:

$x^2 + y^2 + 2gx + 2fy = 0$

Since the circle passes through the point A(2, 0), substitute $x = 2$ and $y = 0$ into the modified equation:

$2^2 + 0^2 + 2g(2) + 2f(0) = 0$

$4 + 4g = 0$

$4g = -4$

Since the circle passes through the point B(0, 2), substitute $x = 0$ and $y = 2$ into the equation $x^2 + y^2 + 2gx + 2fy = 0$:

$0^2 + 2^2 + 2g(0) + 2f(2) = 0$

$4 + 4f = 0$

$4f = -4$

Now, substitute the values of $g = -1$, $f = -1$, and $c = 0$ back into the general equation of the circle $x^2 + y^2 + 2gx + 2fy + c = 0$:

$x^2 + y^2 + 2(-1)x + 2(-1)y + 0 = 0$

$x^2 + y^2 - 2x - 2y = 0$

The equation of the circle passing through the points (2, 0), (0, 2), and (0, 0) is $\mathbf{x^2 + y^2 - 2x - 2y = 0}$.

Given:

Focus F = $(2, 0)$

Directrix equation = $x = -2$ (or $x + 2 = 0$)

To Find:

The equation of the parabola, latus rectum, vertex, and axis.

Solution:

A parabola is defined as the locus of a point P(x, y) that moves such that its distance from a fixed point (the focus) is equal to its distance from a fixed line (the directrix).

Let P be any point $(x, y)$ on the parabola.

The distance of the point P from the focus F(2, 0) is given by the distance formula:

$PF = \sqrt{(x - 2)^2 + (y - 0)^2} = \sqrt{(x - 2)^2 + y^2}$

The distance of the point P(x, y) from the directrix $x + 2 = 0$ is given by the formula for the distance from a point $(x_0, y_0)$ to a line $Ax + By + C = 0$, which is $\frac{|Ax_0 + By_0 + C|}{\sqrt{A^2 + B^2}}$.

Here, $(x_0, y_0) = (x, y)$, $A = 1$, $B = 0$, and $C = 2$.

Distance from P to directrix $= \frac{|1 \cdot x + 0 \cdot y + 2|}{\sqrt{1^2 + 0^2}} = \frac{|x + 2|}{\sqrt{1}} = |x + 2|$

According to the definition of a parabola, the distance from P to the focus is equal to the distance from P to the directrix:

$PF = $ Distance from P to directrix

$\sqrt{(x - 2)^2 + y^2} = |x + 2|$

Square both sides of the equation to eliminate the square root and the absolute value:

$(\sqrt{(x - 2)^2 + y^2})^2 = (|x + 2|)^2$

$(x - 2)^2 + y^2 = (x + 2)^2$

Expand both sides:

$(x^2 - 4x + 4) + y^2 = (x^2 + 4x + 4)$

$x^2 + y^2 - 4x + 4 = x^2 + 4x + 4$

Subtract $x^2 + 4$ from both sides:

$y^2 - 4x = 4x$

Add $4x$ to both sides:

$y^2 = 4x + 4x$

$y^2 = 8x$

The equation of the parabola is $\mathbf{y^2 = 8x}$.

This equation is in the standard form $y^2 = 4ax$, which represents a parabola with vertex at the origin $(0, 0)$, axis along the x-axis, and opening towards the positive x-axis.

Comparing $y^2 = 8x$ with $y^2 = 4ax$, we have:

Dividing by 4, we get $a = 2$.

Now we can find the requested properties:

Vertex: For the standard form $y^2 = 4ax$, the vertex is at $(0, 0)$.

Vertex $= \mathbf{(0, 0)}$.

Axis: For the standard form $y^2 = 4ax$, the axis of the parabola is the x-axis (the line containing the $y=0$).

Axis is the $\mathbf{x-axis}$ (or the line $y=0$).

Latus Rectum: The length of the latus rectum of a parabola of the form $y^2 = 4ax$ is given by $4a$.

Latus Rectum $= 4a = 4(2) = \mathbf{8}$.

We can also verify the focus and directrix from the standard form with $a=2$:

Focus: $(a, 0) = (2, 0)$, which matches the given focus.

Directrix: $x = -a \implies x = -2$, which matches the given directrix.

Summary of results:

Equation of the parabola: $\mathbf{y^2 = 8x}$

Vertex: $\mathbf{(0, 0)}$

Axis: $\mathbf{x-axis}$

Latus Rectum: $\mathbf{8}$

Given:

The equation of the ellipse is $\frac{x^2}{36} + \frac{y^2}{16} = 1$.

To Find:

The lengths of the major and minor axes, the coordinates of the foci, the vertices, the eccentricity, and the latus rectum of the ellipse.

Solution:

The given equation of the ellipse is $\frac{x^2}{36} + \frac{y^2}{16} = 1$.

This equation is in the standard form $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ or $\frac{x^2}{b^2} + \frac{y^2}{a^2} = 1$ for an ellipse centered at the origin $(0, 0)$.

Comparing the denominators, we have $36$ and $16$. Since $36 > 16$, the larger denominator is under the $x^2$ term.

This indicates that the major axis is along the x-axis, and the ellipse is horizontal.

For a horizontal ellipse centered at the origin, the standard form is $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$, where $a^2$ is the larger denominator ($a$ is the semi-major axis) and $b^2$ is the smaller denominator ($b$ is the semi-minor axis).

From the given equation, we have:

$a^2 = 36$

$b^2 = 16$

Taking the square root of $a^2$ and $b^2$ (considering only positive values for $a$ and $b$ as lengths):

$a = \sqrt{36} = 6$

$b = \sqrt{16} = 4$

Length of the major axis:

The length of the major axis is $2a$.

Length of major axis $= 2 \times 6 = \mathbf{12}$.

Length of the minor axis:

The length of the minor axis is $2b$.

Length of minor axis $= 2 \times 4 = \mathbf{8}$.

Coordinates of the vertices:

For a horizontal ellipse centered at the origin, the vertices are located at $(\pm a, 0)$.

Vertices: $(\pm 6, 0)$. The vertices are $\mathbf{(6, 0)}$ and $\mathbf{(-6, 0)}$.

Coordinates of the foci:

To find the foci, we need to find the value of $c$ using the relationship $c^2 = a^2 - b^2$ for an ellipse.

$c^2 = 36 - 16$

$c^2 = 20$

$c = \sqrt{20} = \sqrt{4 \times 5} = \sqrt{4} \times \sqrt{5} = 2\sqrt{5}$

For a horizontal ellipse centered at the origin, the foci are located at $(\pm c, 0)$.

Foci: $(\pm 2\sqrt{5}, 0)$. The foci are $\mathbf{(2\sqrt{5}, 0)}$ and $\mathbf{(-2\sqrt{5}, 0)}$.

Eccentricity:

The eccentricity $e$ of an ellipse is given by $e = \frac{c}{a}$.

$e = \frac{2\sqrt{5}}{6} = \frac{\cancel{2}\sqrt{5}}{\cancel{6}_{3}}$

Eccentricity $e = \mathbf{\frac{\sqrt{5}}{3}}$.

Latus Rectum:

The length of the latus rectum of a horizontal ellipse is given by $\frac{2b^2}{a}$.

Latus Rectum $= \frac{2(16)}{6} = \frac{32}{6} = \frac{\cancel{32}^{16}}{\cancel{6}_{3}}$

Latus Rectum $= \mathbf{\frac{16}{3}}$.

Summary of results:

Length of major axis: $\mathbf{12}$

Length of minor axis: $\mathbf{8}$

Vertices: $\mathbf{(\pm 6, 0)}$

Foci: $\mathbf{(\pm 2\sqrt{5}, 0)}$

Eccentricity: $\mathbf{\frac{\sqrt{5}}{3}}$

Latus Rectum: $\mathbf{\frac{16}{3}}$

Given:

Foci of the hyperbola are $(\pm 5, 0)$.

Length of the latus rectum is $\frac{32}{3}$.

To Find:

The equation of the hyperbola.

Solution:

The foci are given as $(\pm 5, 0)$. Since the foci are on the x-axis and are symmetric with respect to the origin, the center of the hyperbola is at the origin $(0, 0)$ and the transverse axis is along the x-axis.

The standard equation of a horizontal hyperbola centered at the origin is:

$\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$

where $a$ is the distance from the center to a vertex and $b$ is related to the conjugate axis.

From the foci $(\pm 5, 0)$, the distance from the center to a focus is $c = 5$.

So, $c^2 = 5^2 = 25$.

For a horizontal hyperbola, the length of the latus rectum is given by $\frac{2b^2}{a}$.

We are given that the length of the latus rectum is $\frac{32}{3}$.

Multiply both sides by $3a$ and divide by 2:

$6b^2 = 32a$

$3b^2 = 16a$

$b^2 = \frac{16a}{3}$

For a hyperbola, the relationship between $a$, $b$, and $c$ is $c^2 = a^2 + b^2$.

Substitute the values $c^2 = 25$ and $b^2 = \frac{16a}{3}$ into this relationship:

$25 = a^2 + \frac{16a}{3}$

Multiply the entire equation by 3 to eliminate the fraction:

$3 \times 25 = 3 \times a^2 + 3 \times \frac{16a}{3}$

$75 = 3a^2 + 16a$

Rearrange the terms to form a quadratic equation in $a$:

$3a^2 + 16a - 75 = 0$

We can solve this quadratic equation for $a$ using factoring or the quadratic formula. Let's try factoring by splitting the middle term (16a). We look for two numbers that multiply to $3 \times -75 = -225$ and add up to 16. These numbers are 25 and -9.

$3a^2 + 25a - 9a - 75 = 0$

Group the terms and factor:

$a(3a + 25) - 3(3a + 25) = 0$

$(3a + 25)(a - 3) = 0$

This gives two possible values for $a$:

$3a + 25 = 0 \implies 3a = -25 \implies a = -\frac{25}{3}$

$a - 3 = 0 \implies a = 3$

Since $a$ represents the distance from the center to a vertex, it must be a positive value.

Therefore, we take $a = 3$.

Now, we find $a^2$:

$a^2 = 3^2 = 9$

Next, find $b^2$ using the relation $b^2 = \frac{16a}{3}$:

$b^2 = \frac{16(3)}{3} = 16$

Substitute the values of $a^2 = 9$ and $b^2 = 16$ into the standard equation of the horizontal hyperbola $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$:

$\frac{x^2}{9} - \frac{y^2}{16} = 1$

We can verify the given information with this equation:

Vertices: $(\pm a, 0) = (\pm 3, 0)$, consistent with $a=3$.

Foci: $c^2 = a^2 + b^2 = 9 + 16 = 25 \implies c = 5$. Foci are $(\pm c, 0) = (\pm 5, 0)$, consistent with the given foci.

Latus Rectum: $\frac{2b^2}{a} = \frac{2(16)}{3} = \frac{32}{3}$, consistent with the given latus rectum length.

Answer:

The equation of the hyperbola is $\mathbf{\frac{x^2}{9} - \frac{y^2}{16} = 1}$.

Given:

The circle passes through the points A(5, -8), B(2, -9), and C(2, 1).

To Find:

The equation of the circle.

Solution:

Let the general equation of the circle be:

$x^2 + y^2 + 2gx + 2fy + c = 0$

Since the circle passes through the point A(5, -8), substitute $x = 5$ and $y = -8$ into the general equation:

$5^2 + (-8)^2 + 2g(5) + 2f(-8) + c = 0$

$25 + 64 + 10g - 16f + c = 0$

$10g - 16f + c + 89 = 0$}

$10g - 16f + c = -89$

Since the circle passes through the point B(2, -9), substitute $x = 2$ and $y = -9$ into the general equation:

$2^2 + (-9)^2 + 2g(2) + 2f(-9) + c = 0$

$4 + 81 + 4g - 18f + c = 0$}

$4g - 18f + c + 85 = 0$}

$4g - 18f + c = -85$

Since the circle passes through the point C(2, 1), substitute $x = 2$ and $y = 1$ into the general equation:

$2^2 + 1^2 + 2g(2) + 2f(1) + c = 0$

$4 + 1 + 4g + 2f + c = 0$}

$4g + 2f + c + 5 = 0$}

$4g + 2f + c = -5$

We have a system of three linear equations:

(1) $10g - 16f + c = -89$

(2) $4g - 18f + c = -85$

(3) $4g + 2f + c = -5$

Subtract equation (2) from equation (1) to eliminate $c$:

$(10g - 16f + c) - (4g - 18f + c) = -89 - (-85)$

$10g - 4g - 16f + 18f = -89 + 85$

$6g + 2f = -4$

Divide by 2:

$3g + f = -2$

Subtract equation (3) from equation (2) to eliminate $c$:

$(4g - 18f + c) - (4g + 2f + c) = -85 - (-5)$

$4g - 4g - 18f - 2f = -85 + 5$

$-20f = -80$

Divide by -20:

$f = \frac{-80}{-20} = 4$

Substitute the value of $f = 4$ into the equation $3g + f = -2$:

$3g + 4 = -2$

$3g = -2 - 4$

$3g = -6$

$g = -2$

Substitute the values of $g = -2$ and $f = 4$ into equation (3) to find $c$:

$4g + 2f + c = -5$

$4(-2) + 2(4) + c = -5$

$-8 + 8 + c = -5$

$c = -5$

Substitute the values of $g = -2$, $f = 4$, and $c = -5$ into the general equation of the circle $x^2 + y^2 + 2gx + 2fy + c = 0$:

$x^2 + y^2 + 2(-2)x + 2(4)y + (-5) = 0$

$x^2 + y^2 - 4x + 8y - 5 = 0$

The equation of the circle is $\mathbf{x^2 + y^2 - 4x + 8y - 5 = 0}$.

Given:

Vertex V = $(-2, 3)$

Focus F = $(1, 3)$

To Find:

The equation of the parabola, the equation of its directrix, and the length of the latus rectum.

Solution:

The vertex of the parabola is V$(-2, 3)$ and the focus is F$(1, 3)$.

Since the y-coordinates of the vertex and the focus are the same (both are 3), the axis of the parabola is horizontal and is the line passing through the vertex and focus, which is $y = 3$.

The focus $(1, 3)$ is to the right of the vertex $(-2, 3)$ (since $1 > -2$). Therefore, the parabola opens to the right.

The standard equation of a parabola with vertex $(h, k)$ and a horizontal axis opening to the right is:

$(y - k)^2 = 4a(x - h)$

where $(h, k)$ are the coordinates of the vertex and $|a|$ is the distance between the vertex and the focus. Since it opens to the right, $a > 0$.

From the given vertex, we have $(h, k) = (-2, 3)$.

The distance between the vertex V$(-2, 3)$ and the focus F$(1, 3)$ is:

$a = \sqrt{(1 - (-2))^2 + (3 - 3)^2} = \sqrt{(1 + 2)^2 + 0^2} = \sqrt{3^2} = \sqrt{9} = 3$

So, $a = 3$.

Substitute the values of $h = -2$, $k = 3$, and $a = 3$ into the standard equation:

$(y - 3)^2 = 4(3)(x - (-2))$

$(y - 3)^2 = 12(x + 2)$

The equation of the parabola is $\mathbf{(y - 3)^2 = 12(x + 2)}$.

Equation of the Directrix:

For a parabola with vertex $(h, k)$ and a horizontal axis opening to the right, the directrix is a vertical line located at $x = h - a$.

Substitute $h = -2$ and $a = 3$:

$x = -2 - 3$

$x = -5$

The equation of the directrix is $\mathbf{x = -5}$ or $\mathbf{x + 5 = 0}$.

Length of the Latus Rectum:

The length of the latus rectum of a parabola is given by $|4a|$.

Substitute $a = 3$:

Length of latus rectum $= |4(3)| = |12| = \mathbf{12}$.

Summary of results:

Equation of the parabola: $\mathbf{(y - 3)^2 = 12(x + 2)}$

Equation of the directrix: $\mathbf{x = -5}$

Length of the latus rectum: $\mathbf{12}$

Given:

Vertices of the ellipse are $(\pm 6, 0)$.

Foci of the ellipse are $(\pm 4, 0)$.

To Find:

The equation of the ellipse, eccentricity, and the length of the latus rectum.

Solution:

The vertices are given as $(\pm 6, 0)$ and the foci are given as $(\pm 4, 0)$.

Since the vertices and foci are on the x-axis and are symmetric with respect to the origin, the center of the ellipse is at the origin $(0, 0)$ and the major axis is along the x-axis.

The standard equation of a horizontal ellipse centered at the origin is:

$\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$

where $a$ is the length of the semi-major axis and $b$ is the length of the semi-minor axis ($a > b$).

The distance from the center to a vertex is denoted by $a$. From the vertices $(\pm 6, 0)$, we have $a = 6$.

$a^2 = 6^2 = 36$.

The distance from the center to a focus is denoted by $c$. From the foci $(\pm 4, 0)$, we have $c = 4$.

$c^2 = 4^2 = 16$.

For an ellipse, the relationship between $a$, $b$, and $c$ is $c^2 = a^2 - b^2$.

We can use this to find $b^2$:

$b^2 = a^2 - c^2$

Substitute the values $a^2 = 36$ and $c^2 = 16$:

$b^2 = 36 - 16$

$b^2 = 20$

Substitute the values of $a^2 = 36$ and $b^2 = 20$ into the standard equation of the horizontal ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$:

$\frac{x^2}{36} + \frac{y^2}{20} = 1$

The equation of the ellipse is $\mathbf{\frac{x^2}{36} + \frac{y^2}{20} = 1}$.

Eccentricity:

The eccentricity $e$ of an ellipse is given by $e = \frac{c}{a}$.

Substitute the values of $c = 4$ and $a = 6$:

$e = \frac{4}{6} = \frac{\cancel{4}^{2}}{\cancel{6}_{3}}$

Eccentricity $e = \mathbf{\frac{2}{3}}$.

Latus Rectum:

The length of the latus rectum of a horizontal ellipse is given by $\frac{2b^2}{a}$.

Substitute the values $b^2 = 20$ and $a = 6$:

Latus Rectum $= \frac{2(20)}{6} = \frac{40}{6} = \frac{\cancel{40}^{20}}{\cancel{6}_{3}}$

Latus Rectum $= \mathbf{\frac{20}{3}}$.

Summary of results:

Equation of the ellipse: $\mathbf{\frac{x^2}{36} + \frac{y^2}{20} = 1}$

Eccentricity: $\mathbf{\frac{2}{3}}$

Length of the latus rectum: $\mathbf{\frac{20}{3}}$

Given:

Length of the transverse axis is 8.

Foci of the hyperbola are $(\pm 5, 0)$.

To Find:

The equation of the hyperbola.

Solution:

The foci are given as $(\pm 5, 0)$. Since the foci are on the x-axis and are symmetric with respect to the origin, the center of the hyperbola is at the origin $(0, 0)$ and the transverse axis is along the x-axis.

The standard equation of a horizontal hyperbola centered at the origin is:

$\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$

where $a$ is the distance from the center to a vertex (half the length of the transverse axis).

The length of the transverse axis is given as 8. For a horizontal hyperbola centered at the origin, the length of the transverse axis is $2a$.

$2a = 8$

Divide by 2:

$a = 4$

So, $a^2 = 4^2 = 16$.

From the foci $(\pm 5, 0)$, the distance from the center to a focus is $c = 5$.

So, $c^2 = 5^2 = 25$.

For a hyperbola, the relationship between $a$, $b$, and $c$ is $c^2 = a^2 + b^2$.

We can use this relationship to find $b^2$:

$b^2 = c^2 - a^2$

Substitute the values $c^2 = 25$ and $a^2 = 16$:

$b^2 = 25 - 16$

$b^2 = 9$

Now, substitute the values of $a^2 = 16$ and $b^2 = 9$ into the standard equation of the horizontal hyperbola $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$:

$\frac{x^2}{16} - \frac{y^2}{9} = 1$

Answer:

The equation of the hyperbola is $\mathbf{\frac{x^2}{16} - \frac{y^2}{9} = 1}$.

Given:

The circle passes through the points A(1, 1), B(2, -1), and C(3, -2).

To Find:

The equation of the circle.

Solution:

Let the general equation of the circle be:

$x^2 + y^2 + 2gx + 2fy + c = 0$

Since the circle passes through the point A(1, 1), substitute $x = 1$ and $y = 1$ into the general equation:

$1^2 + 1^2 + 2g(1) + 2f(1) + c = 0$

$1 + 1 + 2g + 2f + c = 0$

$2g + 2f + c + 2 = 0$}

$2g + 2f + c = -2$

Since the circle passes through the point B(2, -1), substitute $x = 2$ and $y = -1$ into the general equation:

$2^2 + (-1)^2 + 2g(2) + 2f(-1) + c = 0$

$4 + 1 + 4g - 2f + c = 0$}

$4g - 2f + c + 5 = 0$}

$4g - 2f + c = -5$

Since the circle passes through the point C(3, -2), substitute $x = 3$ and $y = -2$ into the general equation:

$3^2 + (-2)^2 + 2g(3) + 2f(-2) + c = 0$

$9 + 4 + 6g - 4f + c = 0$}

$6g - 4f + c + 13 = 0$}

$6g - 4f + c = -13$

We have a system of three linear equations:

(1) $2g + 2f + c = -2$

(2) $4g - 2f + c = -5$

(3) $6g - 4f + c = -13$

Subtract equation (1) from equation (2) to eliminate $c$:

$(4g - 2f + c) - (2g + 2f + c) = -5 - (-2)$

$4g - 2g - 2f - 2f = -5 + 2$

$2g - 4f = -3$}

Subtract equation (2) from equation (3) to eliminate $c$:

$(6g - 4f + c) - (4g - 2f + c) = -13 - (-5)$

$6g - 4g - 4f + 2f = -13 + 5$

$2g - 2f = -8$}

Divide by 2:

$g - f = -4$

Now we have a system of two linear equations in terms of $g$ and $f$:

(4) $2g - 4f = -3$

(5) $g - f = -4$

From equation (5), express $g$ in terms of $f$: $g = f - 4$.

Substitute this expression for $g$ into equation (4):

$2(f - 4) - 4f = -3$

$2f - 8 - 4f = -3$

$-2f - 8 = -3$

$-2f = -3 + 8$

$-2f = 5$

$f = -\frac{5}{2}$

Substitute the value of $f = -\frac{5}{2}$ back into the expression for $g = f - 4$:

$g = -\frac{5}{2} - 4$

$g = -\frac{5}{2} - \frac{8}{2}$

$g = -\frac{13}{2}$

Substitute the values of $g = -\frac{13}{2}$ and $f = -\frac{5}{2}$ into equation (1) to find $c$:

$2g + 2f + c = -2$

$2\left(-\frac{13}{2}\right) + 2\left(-\frac{5}{2}\right) + c = -2$

$-13 - 5 + c = -2$

$-18 + c = -2$

$c = -2 + 18$

$c = 16$

Substitute the values of $g = -\frac{13}{2}$, $f = -\frac{5}{2}$, and $c = 16$ into the general equation of the circle $x^2 + y^2 + 2gx + 2fy + c = 0$:

$x^2 + y^2 + 2\left(-\frac{13}{2}\right)x + 2\left(-\frac{5}{2}\right)y + 16 = 0$

$x^2 + y^2 - 13x - 5y + 16 = 0$

The equation of the circle is $\mathbf{x^2 + y^2 - 13x - 5y + 16 = 0}$.

Given:

Vertex V = $(0, 0)$ (origin)

The axis of the parabola is the y-axis.

The parabola passes through the point $(-3, 4)$.

To Find:

The equation of the parabola, the focus, and the equation of the directrix.

Solution:

Since the vertex is at the origin $(0, 0)$ and the axis is along the y-axis, the standard equation of the parabola is of the form $x^2 = 4ay$ (opening upwards) or $x^2 = -4ay$ (opening downwards).

The parabola passes through the point $(-3, 4)$. This point has $x = -3$ and $y = 4$. Since the vertex is at $(0,0)$ and the y-coordinate of the point (4) is positive, the parabola must open upwards along the positive y-axis.

The standard equation for a parabola with vertex at the origin and axis along the positive y-axis is:

$x^2 = 4ay$

where $a$ is the distance from the vertex to the focus, and $a > 0$ for the parabola to open upwards.

Substitute the coordinates of the point $(-3, 4)$ into the equation $x^2 = 4ay$:

$(-3)^2 = 4a(4)$

$9 = 16a$}

Solve for $a$:

$a = \frac{9}{16}$

Substitute the value of $a = \frac{9}{16}$ back into the standard equation $x^2 = 4ay$:

$x^2 = 4\left(\frac{9}{16}\right)y$

$x^2 = \frac{36}{16}y$

Simplify the fraction:

$x^2 = \frac{9}{4}y$

The equation of the parabola is $\mathbf{x^2 = \frac{9}{4}y}$ or $\mathbf{4x^2 = 9y}$.

Focus:

For a parabola of the form $x^2 = 4ay$ with vertex at the origin, the focus is located at $(0, a)$.

Using $a = \frac{9}{16}$, the focus is $\mathbf{\left(0, \frac{9}{16}\right)}$.

Equation of the Directrix:

For a parabola of the form $x^2 = 4ay$ with vertex at the origin, the equation of the directrix is $y = -a$.

Using $a = \frac{9}{16}$, the equation of the directrix is $y = -\frac{9}{16}$.

This can also be written as $\mathbf{y + \frac{9}{16} = 0}$ or $\mathbf{16y + 9 = 0}$.

Summary of results:

Equation of the parabola: $\mathbf{x^2 = \frac{9}{4}y}$

Focus: $\mathbf{\left(0, \frac{9}{16}\right)}$

Equation of the directrix: $\mathbf{y = -\frac{9}{16}}$

Given: Foci $(\pm \sqrt{5}, 0)$, Latus Rectum length $\frac{16}{3}$.

To Find: Equation of the ellipse.

Solution:

Since foci are $(\pm \sqrt{5}, 0)$, the center is at the origin $(0, 0)$, the major axis is along the x-axis, and $c = \sqrt{5}$. Thus $c^2 = 5$.

The standard equation is $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$, where $a > b$.

The length of the latus rectum is $\frac{2b^2}{a}$. Given $\frac{2b^2}{a} = \frac{16}{3}$, we get $6b^2 = 16a$, so $b^2 = \frac{8a}{3}$.

For an ellipse, $c^2 = a^2 - b^2$. Substitute the values of $c^2$ and $b^2$:

$5 = a^2 - \frac{8a}{3}$

Multiply by 3: $15 = 3a^2 - 8a$.

Rearrange: $3a^2 - 8a - 15 = 0$.

Using the quadratic formula $a = \frac{-B \pm \sqrt{B^2 - 4AC}}{2A}$ for $3a^2 - 8a - 15 = 0$ ($A=3, B=-8, C=-15$):

$a = \frac{8 \pm \sqrt{(-8)^2 - 4(3)(-15)}}{2(3)} = \frac{8 \pm \sqrt{64 + 180}}{6} = \frac{8 \pm \sqrt{244}}{6} = \frac{8 \pm 2\sqrt{61}}{6} = \frac{4 \pm \sqrt{61}}{3}$.

Since $a$ must be positive and greater than $c = \sqrt{5} \approx 2.236$, we choose $a = \frac{4 + \sqrt{61}}{3}$ because $\frac{4 + \sqrt{61}}{3} \approx \frac{4 + 7.81}{3} \approx 3.937$, which is positive and greater than $\sqrt{5}$. The other root $\frac{4 - \sqrt{61}}{3}$ is negative.

Calculate $a^2$: $a^2 = \left(\frac{4 + \sqrt{61}}{3}\right)^2 = \frac{16 + 8\sqrt{61} + 61}{9} = \frac{77 + 8\sqrt{61}}{9}$.

Calculate $b^2$ using $b^2 = \frac{8a}{3}$: $b^2 = \frac{8}{3} \left(\frac{4 + \sqrt{61}}{3}\right) = \frac{32 + 8\sqrt{61}}{9}$.

Substitute $a^2$ and $b^2$ into the ellipse equation $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$:

$\frac{x^2}{\frac{77 + 8\sqrt{61}}{9}} + \frac{y^2}{\frac{32 + 8\sqrt{61}}{9}} = 1$}

$\frac{9x^2}{77 + 8\sqrt{61}} + \frac{9y^2}{32 + 8\sqrt{61}} = 1$}

Answer: The equation of the ellipse is $\mathbf{\frac{9x^2}{77 + 8\sqrt{61}} + \frac{9y^2}{32 + 8\sqrt{61}} = 1}$.

Given:

The equation of the hyperbola is $\frac{x^2}{16} - \frac{y^2}{9} = 1$.

To Find:

The equations of the asymptotes, the eccentricity, and the length of the conjugate axis.

Solution:

The given equation of the hyperbola is $\frac{x^2}{16} - \frac{y^2}{9} = 1$.

This equation is in the standard form of a hyperbola centered at the origin $(0, 0)$ with a horizontal transverse axis:

$\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$

Comparing the given equation with the standard form, we have:

$a^2 = 16 \implies a = \sqrt{16} = 4$

$b^2 = 9 \implies b = \sqrt{9} = 3$

Equations of the Asymptotes:

For a horizontal hyperbola centered at the origin, the equations of the asymptotes are given by $y = \pm \frac{b}{a}x$.

Substitute the values of $a=4$ and $b=3$:

$y = \pm \frac{3}{4}x$

The equations of the asymptotes are $\mathbf{y = \frac{3}{4}x}$ and $\mathbf{y = -\frac{3}{4}x}$.

Eccentricity:

To find the eccentricity $e$, we first need to find the value of $c$ using the relationship $c^2 = a^2 + b^2$ for a hyperbola.

$c^2 = 16 + 9$

$c^2 = 25$

$c = \sqrt{25} = 5$

The eccentricity $e$ of a hyperbola is defined as the ratio $e = \frac{c}{a}$.

Substitute the values of $c = 5$ and $a = 4$:

$e = \frac{5}{4}$

Eccentricity $e = \mathbf{\frac{5}{4}}$.

Length of the Conjugate Axis:

For a hyperbola, the length of the conjugate axis is $2b$.

Substitute the value of $b=3$:

Length of conjugate axis $= 2 \times 3 = \mathbf{6}$.

Summary of results:

Equations of the asymptotes: $\mathbf{y = \pm \frac{3}{4}x}$

Eccentricity: $\mathbf{\frac{5}{4}}$

Length of the conjugate axis: $\mathbf{6}$