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Class 11th Chapters
1. Sets 2. Relations and Functions 3. Trigonometric Functions
4. Principle of Mathematical Induction 5. Complex Numbers and Quadratic Equations 6. Linear Inequalities
7. Permutations and Combinations 8. Binomial Theorem 9. Sequences and Series
10. Straight Lines 11. Conic Sections 12. Introduction to Three Dimensional Geometry
13. Limits and Derivatives 14. Mathematical Reasoning 15. Statistics
16. Probability

Content On This Page
Sections of a Cone Circle and its Various Forms Geometrical Condition for the Intersection of a Line and a Circle
Relative Position of Two CIrcles Symmetry Parabola
Ellipse Hyperbola Parametric Equations of Conics
Applications


Chapter 11 Conic Sections (Concepts)

Welcome to an illuminating exploration of Conic Sections, a fascinating family of curves that have captivated mathematicians, astronomers, and physicists for millennia. This chapter delves into the geometry of these shapes, which arise naturally from the intersection of a plane with a double-napped right circular cone. Imagine slicing through a cone at different angles; the resulting edge of the cut reveals these elegant curves – the circle, the parabola, the ellipse, and the hyperbola. Alternatively, and perhaps more profoundly, each conic section can be defined analytically as a locus of points in a plane. This locus definition involves a fixed point (the focus), a fixed line (the directrix), and a constant positive ratio known as the eccentricity, denoted by $e$. A conic section is the set of all points P such that the ratio of the distance from P to the focus to the perpendicular distance from P to the directrix is equal to $e$. The specific value of this eccentricity $e$ uniquely determines the type of curve generated.

Our study begins, perhaps counter-intuitively, with the simplest case, often considered a degenerate form of the ellipse: the Circle ($e=0$). While circles were studied earlier, we revisit them here within the conic context. The standard algebraic equation for a circle centered at $(h, k)$ with radius $r$ is given by the familiar formula $\mathbf{(x - h)^2 + (y - k)^2 = r^2}$. We also examine the general form of the circle's equation, $x^2 + y^2 + 2gx + 2fy + c = 0$, learning how to extract its center $(-g, -f)$ and radius $\sqrt{g^2 + f^2 - c}$.

Next, we investigate the Parabola, defined by an eccentricity $e=1$. This means a parabola is the locus of all points that are equidistant from the focus and the directrix. We explore the four standard orientations of parabolas whose vertex is at the origin: $y^2 = 4ax$, $y^2 = -4ax$, $x^2 = 4ay$, and $x^2 = -4ay$. For each standard form, we learn to identify its key components: the coordinates of the focus, the equation of the directrix, the axis of symmetry, the location of the vertex (at the origin for these standard forms), and the length of the latus rectum (the chord through the focus perpendicular to the axis, with length $4a$).

We then move to the Ellipse, characterized by an eccentricity $0 < e < 1$. Geometrically, an ellipse is the locus of points such that the sum of the distances from any point on the ellipse to two fixed points (the foci, plural of focus) is constant. We analyze the two standard forms of the ellipse centered at the origin: $\mathbf{\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1}$ and $\mathbf{\frac{x^2}{b^2} + \frac{y^2}{a^2} = 1}$ (where $a > b > 0$). For each, we determine the coordinates of the foci, the vertices, the lengths of the major axis ($2a$) and minor axis ($2b$), the equations of the directrices, the eccentricity ($e = \frac{c}{a}$, where $c^2 = a^2 - b^2$), and the length of the latus rectum ($\frac{2b^2}{a}$).

Finally, we explore the Hyperbola, defined by an eccentricity $e > 1$. A hyperbola is the locus of points where the absolute difference of the distances from any point on the hyperbola to two fixed points (the foci) is constant. We study its two standard forms centered at the origin: $\mathbf{\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1}$ (transverse axis along x-axis) and $\mathbf{\frac{y^2}{a^2} - \frac{x^2}{b^2} = 1}$ (transverse axis along y-axis). Similar to the ellipse, we identify the coordinates of its foci and vertices, the lengths of the transverse axis ($2a$) and conjugate axis ($2b$), the equations of the directrices, the eccentricity ($e = \frac{c}{a}$, where now $c^2 = a^2 + b^2$), and the length of the latus rectum ($\frac{2b^2}{a}$). The concept of asymptotes – lines that the hyperbola approaches but never touches as it extends to infinity – is also often introduced. Throughout this chapter, a key skill is identifying the type of conic section represented by a given quadratic equation (often requiring algebraic manipulation like completing the square to reach a standard form) and subsequently determining all its characteristic parameters and sketching its graph accurately.



Sections of a Cone

The curves that are the focus of this chapter—the circle, ellipse, parabola, and hyperbola—are collectively known as conic sections or simply conics. Their study is a cornerstone of geometry and has profound applications in fields like astronomy (planetary orbits), physics (parabolic reflectors), and engineering (hyperbolic cooling towers).

They are named "conic sections" because each one can be generated by the intersection of a flat, two-dimensional plane and a three-dimensional double-napped cone. The study of these beautiful and important curves dates back to the ancient Greek mathematicians, notably Apollonius of Perga, who first systematically described them.


Fundamental Definitions

To understand how conic sections are formed, we must first define the three-dimensional object being sliced: a double-napped right circular cone.

The Double-Napped Cone

A cone is a surface generated by a moving line. To construct it, we need a few key elements:

A double-napped right circular cone is the surface formed when we rotate the generator line (g) 360 degrees around the axis, while keeping the vertex and the semi-vertical angle fixed.

Diagram showing a double-napped right circular cone with its axis, vertex, generator, and the semi-vertical angle Beta labelled.

The Intersecting Plane

A conic section is the curve formed by the intersection of this cone with a plane (a flat, two-dimensional surface). The specific type of curve we get depends entirely on the orientation of this plane relative to the cone.

We define this orientation using a second angle, $\alpha$:

The entire classification of conic sections is based on the comparison between the plane's angle, $\alpha$, and the cone's semi-vertical angle, $\beta$.


Types of Conic Sections (Non-Degenerate Cases)

When the slicing plane does not pass through the vertex of the cone, we get one of the four famous conic sections. The specific curve is determined by the relationship between $\alpha$ and $\beta$.

1. Circle ($\alpha = 90^\circ$)

When the cutting plane is perpendicular to the axis of the cone, it cuts straight across one nappe. In this case, the angle $\alpha$ is exactly $90^\circ$.

The resulting cross-section is a perfect circle. A circle can be considered a special, highly symmetric case of an ellipse.

A plane cutting a single nappe of a cone perpendicular to the axis (alpha = 90 degrees), forming a circle.

2. Ellipse ($\beta < \alpha < 90^\circ$)

When the plane is tilted so it is no longer perpendicular to the axis, but its angle $\alpha$ is still greater than the semi-vertical angle $\beta$, it cuts completely through one nappe on a slant.

This creates a closed, oval-shaped curve called an ellipse. All circles are ellipses, but not all ellipses are circles.

A plane cutting a single nappe of a cone at an angle Alpha such that Beta < Alpha < 90 degrees, forming an ellipse.

3. Parabola ($\alpha = \beta$)

When the plane is tilted such that it is exactly parallel to a generator line of the cone, its angle $\alpha$ becomes equal to the semi-vertical angle $\beta$.

The plane intersects only one nappe, forming an open curve that extends infinitely and never closes. This U-shaped curve is a parabola. It represents the boundary case between the closed ellipse and the two-branched hyperbola.

A plane cutting a single nappe of a cone parallel to the generator line (Alpha = Beta), forming a parabola.

4. Hyperbola ($0 \le \alpha < \beta$)

When the plane is tilted even further, so that it is steeper than the generator (meaning its angle $\alpha$ is less than the semi-vertical angle $\beta$), it is guaranteed to intersect both nappes of the cone.

This creates two separate, open curves that are mirror images of each other. This two-branched curve is a hyperbola.

A plane cutting both nappes of a cone (Alpha < Beta), forming a hyperbola with two branches.

Degenerate Conic Sections

A special situation occurs if the slicing plane passes exactly through the vertex of the cone. In this case, the intersection is called a degenerate conic. The term "degenerate" is used in mathematics to refer to a limiting case where a class of objects changes its nature and becomes simpler.

These are not curves in the usual sense but rather simpler geometric figures like points and lines. The type of degenerate conic formed still depends on the relationship between $\alpha$ and $\beta$.

A composite image showing the three degenerate cases: a point, a single line, and a pair of intersecting lines, all formed by a plane passing through the cone's vertex.

While the conic sections are elegantly defined by slicing a cone, our study will focus on their properties and equations as they are described within the two-dimensional Cartesian plane.



Circle and its Various Forms

The circle is one of the four non-degenerate conic sections and is arguably the most familiar. It is defined by a simple and elegant geometric property.

Geometrically, a circle is the set (or locus) of all points in a plane that are at a fixed, constant distance from a fixed point. The fixed point is called the center, and the fixed distance is called the radius.

A circle with its center C and radius r labelled. A point P on the circumference is shown, with the distance CP being equal to r.

1. Standard Equation of a Circle (Center-Radius Form)

We can derive the algebraic equation of a circle directly from its definition by using the distance formula. This form is the most intuitive as it directly relates to the geometric definition.

Derivation

Let the center of the circle be the point $C(h, k)$, and let the radius be a constant value $r$, where $r > 0$.

Let $P(x, y)$ be any point on the circumference of the circle. By definition, the distance from the center C to any such point P must be equal to the radius $r$.

Distance(C, P) = $r$

A circle on the Cartesian plane with center C(h,k) and a point P(x,y) on the circumference. A right-angled triangle is shown with sides (x-h) and (y-k) and hypotenuse r.

Using the distance formula between two points, we have:

$\sqrt{(x-h)^2 + (y-k)^2} = r$

To eliminate the square root and obtain a simpler polynomial form, we square both sides of the equation.

$(\sqrt{(x-h)^2 + (y-k)^2})^2 = r^2$

This gives us the standard equation of a circle, also known as the center-radius form:

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

... (i)

This form is very useful because the coordinates of the center $(h, k)$ and the square of the radius $r^2$ are explicitly visible in the equation.


2. Central Form (Circle Centered at the Origin)

A common special case of the standard form is when a circle's center is located at the origin of the Cartesian plane, i.e., at the point $(0, 0)$.

Derivation

In this case, the values for the center are $h=0$ and $k=0$. Substituting these into the standard equation gives:

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

This simplifies to the central form:

$x^2 + y^2 = r^2$

... (ii)

A circle centered at the origin (0,0) with radius r.

3. General Equation of a Circle

If we expand the standard equation of a circle and rearrange the terms, we arrive at a more general form of the equation.

Derivation

Starting with the standard form, $(x-h)^2 + (y-k)^2 = r^2$:

Expand the squared binomials:

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

Group the terms by degree and move all terms to the left side:

$x^2 + y^2 - 2hx - 2ky + (h^2 + k^2 - r^2) = 0$

This equation is a second-degree equation in $x$ and $y$. To simplify it, we introduce new constants. Let $g = -h$, $f = -k$, and $c = h^2 + k^2 - r^2$. Substituting these gives:

$x^2 + y^2 + 2(-h)x + 2(-k)y + c = 0$

The general equation of a circle is:

$x^2 + y^2 + 2gx + 2fy + c = 0$

... (iii)

From this general form, we can find the center and radius by reversing the substitutions:

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

... (iv)

Key Points for the General Equation

For a general second-degree equation $ax^2 + 2hxy + by^2 + 2gx + 2fy + c = 0$ to represent a circle, it must satisfy two conditions:

  1. The coefficients of $x^2$ and $y^2$ must be equal ($a=b$). We usually divide the equation to make them both 1.
  2. There must be no $xy$ term (the coefficient $h$ must be 0).

Nature of the Circle from the General Equation

The value of the expression under the square root, $g^2 + f^2 - c$, determines what the equation represents:


4. Diameter Form of a Circle

This form is used when the coordinates of the endpoints of a diameter are known.

Derivation

Let the endpoints of a diameter be $A(x_1, y_1)$ and $B(x_2, y_2)$.

A fundamental property of a circle is that the angle subtended by a diameter at any point on the circumference is a right angle ($90^\circ$).

Let $P(x, y)$ be any point on the circle. Then the line segment PA must be perpendicular to the line segment PB.

A circle with diameter AB. A point P on the circumference forms a right angle at APB.

The condition for two lines to be perpendicular is that the product of their slopes is -1.

Slope of PA, $m_1 = \frac{y - y_1}{x - x_1}$

Slope of PB, $m_2 = \frac{y - y_2}{x - x_2}$

Setting the product of the slopes to -1:

$m_1 \times m_2 = -1$

$\left(\frac{y - y_1}{x - x_1}\right) \left(\frac{y - y_2}{x - x_2}\right) = -1$

$(y - y_1)(y - y_2) = -1 \times (x - x_1)(x - x_2)$

Moving all terms to one side gives the diameter form of the equation of a circle:

$(x - x_1)(x - x_2) + (y - y_1)(y - y_2) = 0$

... (v)


Example 1. Find the equation of the circle with center (2, -3) and radius 4.

Answer:

Given: Center $(h, k) = (2, -3)$ and Radius $r = 4$.

Solution:

We use the standard center-radius form: $(x-h)^2 + (y-k)^2 = r^2$.

Substitute the given values: $(x - 2)^2 + (y - (-3))^2 = (4)^2$.

This gives the standard form: $(x - 2)^2 + (y + 3)^2 = 16$.

To find the general form, we expand: $(x^2 - 4x + 4) + (y^2 + 6y + 9) = 16$.

Combining terms gives the general form: $x^2 + y^2 - 4x + 6y - 3 = 0$.


Example 2. Find the center and radius of the circle $x^2 + y^2 + 6x - 10y + 18 = 0$.

Answer:

Given: The equation is $x^2 + y^2 + 6x - 10y + 18 = 0$.

Solution (Completing the Square):

Group the x-terms and y-terms: $(x^2 + 6x) + (y^2 - 10y) = -18$.

Complete the square for x: add $(6/2)^2 = 9$ to both sides.

Complete the square for y: add $(-10/2)^2 = 25$ to both sides.

$(x^2 + 6x + 9) + (y^2 - 10y + 25) = -18 + 9 + 25$

Factor the perfect squares: $(x + 3)^2 + (y - 5)^2 = 16$.

Comparing this with $(x - h)^2 + (y - k)^2 = r^2$, we get:

Center $(h, k) = (-3, 5)$ and radius $r^2 = 16 \implies r = 4$.

The final answer is Center = (-3, 5), Radius = 4.


Example 3. Find the equation of the circle whose diameter has endpoints at A(1, 4) and B(5, 2).

Answer:

Given:

Endpoints of the diameter are $(x_1, y_1) = (1, 4)$ and $(x_2, y_2) = (5, 2)$.

Solution:

We use the diameter form of the circle's equation:

$(x - x_1)(x - x_2) + (y - y_1)(y - y_2) = 0$

Substitute the coordinates of the endpoints:

$(x - 1)(x - 5) + (y - 4)(y - 2) = 0$

Expand the products:

$(x^2 - 5x - x + 5) + (y^2 - 2y - 4y + 8) = 0$

$(x^2 - 6x + 5) + (y^2 - 6y + 8) = 0$

Combine like terms to get the general form:

$x^2 + y^2 - 6x - 6y + 13 = 0$

The required equation is $x^2 + y^2 - 6x - 6y + 13 = 0$.



Geometrical Condition for the Intersection of a Line and a Circle

When we consider a straight line and a circle together in a plane, there are three distinct possibilities for how they can interact with each other:

  1. The line can pass through the circle, cutting it at two distinct points.
  2. The line can just touch the circle at a single point.
  3. The line can miss the circle entirely.

We can determine which of these scenarios occurs by using a simple geometric relationship involving the circle's center, its radius, and the line.


Condition Based on Distance from Center to Line

The key to determining the nature of the intersection is to calculate the perpendicular distance from the center of the circle to the line and compare this distance with the radius of the circle.

Let the circle have a center at $C(h, k)$ and a radius of $r$.

Let the line have the general equation $Ax + By + C = 0$.

First, we calculate the perpendicular distance, $d$, from the center $C(h, k)$ to the line using the formula:

$d = \frac{|Ah + Bk + C|}{\sqrt{A^2 + B^2}}$

Once we have this distance $d$, we compare it to the radius $r$.

1. The Line is a Secant (Two Intersection Points)

If the distance from the center to the line is less than the radius, the line must pass through the interior of the circle. This means it will intersect the circumference at two distinct points. Such a line is called a secant.

Condition: $d < r$

A circle with center C and radius r. A line passes through the circle, with the perpendicular distance d from C to the line shown to be shorter than r.

2. The Line is a Tangent (One Intersection Point)

If the distance from the center to the line is exactly equal to the radius, the line just touches the circle at a single point. This point is called the point of tangency, and the line is a tangent.

Condition: $d = r$

A circle with center C and radius r. A line is shown tangent to the circle, with the perpendicular distance d from C to the line shown to be equal to r.

3. The Line and Circle Do Not Intersect

If the distance from the center to the line is greater than the radius, the line is too far away from the center to touch the circle. It passes entirely outside the circle, and there are no intersection points.

Condition: $d > r$

A circle with center C and radius r. A line is shown outside the circle, with the perpendicular distance d from C to the line shown to be longer than r.

Example 1. Determine whether the line $x + y = 4$ intersects, is tangent to, or does not intersect the circle $x^2 + y^2 = 4$.

Answer:

Given:

Circle: $x^2 + y^2 = 4$.

Line: $x + y = 4$.

Solution:

Step 1: Find the center and radius of the circle.

The equation $x^2 + y^2 = 4$ is in the form $x^2 + y^2 = r^2$.

Center $C(h, k) = (0, 0)$.

Radius $r = \sqrt{4} = 2$.

Step 2: Find the perpendicular distance from the center to the line.

First, write the line's equation in the general form $Ax + By + C = 0$.

$x + y - 4 = 0$. So, $A=1, B=1, C=-4$.

Now, calculate the distance $d$ from the center $(0, 0)$ to this line.

$d = \frac{|A(0) + B(0) + C|}{\sqrt{A^2 + B^2}} = \frac{|C|}{\sqrt{A^2+B^2}}$

$d = \frac{|-4|}{\sqrt{1^2 + 1^2}} = \frac{4}{\sqrt{2}}$

Rationalizing the denominator, we get:

$d = \frac{4\sqrt{2}}{2} = 2\sqrt{2}$

Step 3: Compare the distance $d$ with the radius $r$.

We have $r = 2$ and $d = 2\sqrt{2}$.

Since $\sqrt{2} \approx 1.414$, the distance is $d \approx 2(1.414) = 2.828$.

Clearly, $2.828 > 2$, so $d > r$.

Because the distance from the center to the line is greater than the radius, the line does not intersect the circle.


Example 2. For what value(s) of $k$ will the line $3x + 4y = k$ be tangent to the circle $x^2 + y^2 = 25$?

Answer:

Given:

Circle: $x^2 + y^2 = 25$.

Line: $3x + 4y = k$.

Solution:

Step 1: Find the center and radius of the circle.

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

Center $C(h, k) = (0, 0)$.

Radius $r = \sqrt{25} = 5$.

Step 2: Set up the condition for tangency.

For the line to be tangent to the circle, the perpendicular distance $d$ from the center to the line must be equal to the radius $r$.

$d = r$

Step 3: Calculate the distance and form an equation.

Write the line's equation in the general form: $3x + 4y - k = 0$.

Here, $A=3, B=4,$ and $C=-k$.

The distance from the center $(0, 0)$ to this line is:

$d = \frac{|A(0) + B(0) + C|}{\sqrt{A^2 + B^2}} = \frac{|-k|}{\sqrt{3^2 + 4^2}} = \frac{|k|}{\sqrt{9+16}} = \frac{|k|}{\sqrt{25}} = \frac{|k|}{5}$

Step 4: Solve the equation for $k$.

Now, we apply the tangency condition, $d = r$.

$\frac{|k|}{5} = 5$

Multiply both sides by 5.

$|k| = 25$

This absolute value equation gives two possible solutions for $k$.

$k = 25$ or $k = -25$.

The values of $k$ for which the line is tangent to the circle are $k = \pm 25$.



Relative Position of Two Circles

When two circles are placed in the same plane, their relationship can be fully described by comparing the distance between their centers to their respective radii. This geometric analysis allows us to determine, without solving any simultaneous equations, whether the circles are separate from each other, touch at a single point, or intersect at two distinct points. This method provides a powerful visual and analytical tool for understanding the geometry of circles.


The Key Comparison: Distance vs. Radii

To analyze the position of two circles, we first need to establish three fundamental quantities:

The entire relationship between the two circles hinges on comparing the distance $d$ with two critical values: the sum of the radii ($r_1 + r_2$) and the absolute difference of the radii ($|r_1 - r_2|$). Let's explore the possible cases that arise from this comparison.

1. One Circle is Completely Outside the Other

This occurs when the two circles are completely separate, with a gap between them. For this to happen, the distance between their centers must be larger than the distance required to just make them touch, which is the sum of their radii.

Condition: $d > r_1 + r_2$

Two circles are completely outside each other. The distance d between their centers is clearly greater than the sum of their radii, r1 + r2.

2. Circles Touch Externally

In this case, the two circles touch at a single point, but neither is inside the other. The centers and the point of contact lie on a straight line. The distance between the centers is precisely the length of the first radius plus the length of the second radius.

Condition: $d = r_1 + r_2$

Two circles touching externally at one point. The distance d between their centers is exactly equal to the sum of their radii, r1 + r2.

3. Circles Intersect at Two Distinct Points

For the circles to overlap and cross each other, the distance between their centers must be small enough for them to reach each other ($d < r_1 + r_2$), but large enough that one circle is not completely contained within the other ($d > |r_1 - r_2|$). This range of distances allows for two points of intersection.

Condition: $|r_1 - r_2| < d < r_1 + r_2$

Two circles intersecting at two distinct points. The distance d between their centers is greater than the difference of their radii but less than their sum.

4. Circles Touch Internally

This occurs when one circle is inside the other and they touch at a single point. Imagine the larger circle with radius $r_1$ and center $C_1$. For the smaller circle (radius $r_2$) to touch its inner edge, the distance from $C_1$ to $C_2$ ($d$) plus the radius of the smaller circle ($r_2$) must equal the radius of the larger circle ($r_1$). So, $d + r_2 = r_1$, which rearranges to $d = r_1 - r_2$. The absolute value accounts for either circle being the larger one.

Condition: $d = |r_1 - r_2|$

Two circles touching internally at one point. The distance d between their centers is exactly equal to the absolute difference of their radii, |r1 - r2|.

5. One Circle is Completely Inside the Other (Without Touching)

If one circle is inside the other and there is a gap between their circumferences, the distance between their centers must be even smaller than the distance required for them to touch internally. This means the distance $d$ is less than the absolute difference of their radii.

Condition: $d < |r_1 - r_2|$

One smaller circle is completely inside a larger one, not touching. The distance d between their centers is less than the absolute difference of their radii.

Special Case: Concentric Circles

If the distance between the centers is zero, the circles share the exact same center. If their radii are different, they form concentric rings. If their radii are the same, they are the same circle.

Condition: $d = 0$


Example 1. Determine the relative position of the circles $x^2 + y^2 - 4x + 2y - 4 = 0$ and $x^2 + y^2 + 6x - 8y + 24 = 0$.

Answer:

Given:

Circle 1: $x^2 + y^2 - 4x + 2y - 4 = 0$

Circle 2: $x^2 + y^2 + 6x - 8y + 24 = 0$

Solution:

Step 1: Find the center and radius of each circle.

We compare each equation with the general form $x^2 + y^2 + 2gx + 2fy + c = 0$.

For Circle 1:

$2g_1 = -4 \implies g_1 = -2$

$2f_1 = 2 \implies f_1 = 1$

$c_1 = -4$

Center $C_1 = (-g_1, -f_1) = (2, -1)$.

Radius $r_1 = \sqrt{g_1^2 + f_1^2 - c_1} = \sqrt{(-2)^2 + 1^2 - (-4)} = \sqrt{4 + 1 + 4} = \sqrt{9} = 3$.

For Circle 2:

$2g_2 = 6 \implies g_2 = 3$

$2f_2 = -8 \implies f_2 = -4$

$c_2 = 24$

Center $C_2 = (-g_2, -f_2) = (-3, 4)$.

Radius $r_2 = \sqrt{g_2^2 + f_2^2 - c_2} = \sqrt{3^2 + (-4)^2 - 24} = \sqrt{9 + 16 - 24} = \sqrt{1} = 1$.

Step 2: Calculate the distance between the centers.

The distance $d$ between $C_1(2, -1)$ and $C_2(-3, 4)$ is:

$d = \sqrt{(-3 - 2)^2 + (4 - (-1))^2}$

$d = \sqrt{(-5)^2 + (5)^2} = \sqrt{25 + 25} = \sqrt{50}$

$d = 5\sqrt{2}$

Step 3: Compare the distance with the sum and difference of the radii.

The values we need to compare are:

  • Distance between centers: $d = 5\sqrt{2} \approx 7.07$
  • Sum of radii: $r_1 + r_2 = 3 + 1 = 4$
  • Absolute difference of radii: $|r_1 - r_2| = |3 - 1| = 2$

We observe that $7.07 > 4$.

Therefore, the distance between the centers is greater than the sum of the radii ($d > r_1 + r_2$).

Based on this condition, the two circles lie completely outside each other without touching.



Symmetry

Symmetry is a fundamental concept in geometry that describes a sense of balance, harmony, or self-similarity in a shape. A figure is symmetric if it remains unchanged after undergoing a certain transformation, such as reflection, rotation, or translation.

In coordinate geometry, we can use the algebraic equation of a curve to precisely test for various types of symmetry. Understanding a curve's symmetry is a powerful tool that helps us visualize and sketch its graph more easily, as it often means we only need to plot points for a portion of the curve and can then infer the rest.


Tests for Symmetry

For a curve given by an equation, say $F(x, y) = 0$, we can determine if it possesses symmetry with respect to the coordinate axes or the origin by performing simple algebraic tests.

These tests are based on the core idea that if a curve is symmetric with respect to a line or a point, then for every point $(x, y)$ on the curve, its reflection across that line or through that point must also be a point on the curve, thereby satisfying its equation.

1. Symmetry with respect to the x-axis (Horizontal Symmetry)

A curve is symmetric about the x-axis if the part of the curve above the x-axis is a mirror image of the part below it. This means that if a point $(x, y)$ is on the curve, its reflection across the x-axis, the point $(x, -y)$, must also be on the curve.

Algebraic Test: The equation of the curve remains unchanged when $y$ is replaced by $-y$. This is often true for equations where $y$ only appears with even powers (e.g., $y^2, y^4$).

A curve that is symmetric about the x-axis. For a point (x,y) on the curve, the point (x,-y) is also on the curve.

2. Symmetry with respect to the y-axis (Vertical Symmetry)

A curve is symmetric about the y-axis if the part of the curve to the right of the y-axis is a mirror image of the part to the left. This means that if a point $(x, y)$ is on the curve, its reflection across the y-axis, the point $(-x, y)$, must also be on the curve.

Algebraic Test: The equation of the curve remains unchanged when $x$ is replaced by $-x$. This is often true for equations where $x$ only appears with even powers (e.g., $x^2, x^4$).

A curve that is symmetric about the y-axis. For a point (x,y) on the curve, the point (-x,y) is also on the curve.

3. Symmetry with respect to the Origin (Point Symmetry)

A curve is symmetric about the origin if for any point $(x, y)$ on the curve, the point $(-x, -y)$ is also on the curve. Geometrically, this means that if you draw a line segment from $(x, y)$ through the origin, it will meet the curve again at $(-x, -y)$, and the origin will be the midpoint of this segment.

This is equivalent to a 180-degree rotational symmetry about the origin. If you rotate the entire graph 180 degrees around the origin, it will look identical to how it started.

Algebraic Test: The equation of the curve remains unchanged when $x$ is replaced by $-x$ and $y$ is replaced by $-y$ simultaneously.

A curve that is symmetric about the origin. For a point (x,y) on the curve, the point (-x,-y) is also on the curve.

4. Symmetry with respect to the Line $y=x$

A curve is symmetric about the line $y=x$ if for any point $(x, y)$ on the curve, the point $(y, x)$ is also on the curve. This type of symmetry is of special importance when studying inverse functions, as the graph of a function and its inverse are always reflections of each other across the line $y=x$.

Algebraic Test: The equation of the curve remains unchanged when $x$ and $y$ are interchanged.

A curve and its reflection are shown to be symmetric with respect to the dashed line y=x. For a point (x,y) on the curve, the point (y,x) is on the reflected curve.

Symmetry of Standard Conic Sections

The standard conic sections, when centered at the origin or with an axis on a coordinate axis, exhibit clear symmetries that can be verified with the algebraic tests:


Example 1. Test the equation $y^2 = 4x + 8$ for symmetry with respect to the x-axis, y-axis, and the origin.

Answer:

Given: The equation of the curve is $y^2 = 4x + 8$.

Solution:

1. Test for x-axis symmetry: Replace $y$ with $-y$.

$(-y)^2 = 4x + 8 \implies y^2 = 4x + 8$.

The equation is unchanged. Symmetric about the x-axis.

2. Test for y-axis symmetry: Replace $x$ with $-x$.

$y^2 = 4(-x) + 8 \implies y^2 = -4x + 8$.

The equation is different. Not symmetric about the y-axis.

3. Test for origin symmetry: Replace $x$ with $-x$ and $y$ with $-y$.

$(-y)^2 = 4(-x) + 8 \implies y^2 = -4x + 8$.

The equation is different. Not symmetric about the origin.


Example 2. Test the equation $x^2 + y^2 = 25$ for all four types of symmetry.

Answer:

Given: The equation of the curve is $x^2 + y^2 = 25$.

Solution:

1. x-axis symmetry (replace y with -y): $x^2 + (-y)^2 = 25 \implies x^2 + y^2 = 25$. Unchanged. Symmetric.

2. y-axis symmetry (replace x with -x): $(-x)^2 + y^2 = 25 \implies x^2 + y^2 = 25$. Unchanged. Symmetric.

3. Origin symmetry (replace x with -x, y with -y): $(-x)^2 + (-y)^2 = 25 \implies x^2 + y^2 = 25$. Unchanged. Symmetric.

4. Line $y=x$ symmetry (interchange x and y): $y^2 + x^2 = 25 \implies x^2 + y^2 = 25$. Unchanged. Symmetric.


Example 3. Test the equation $xy = 4$ for symmetry.

Answer:

Given: The equation of the curve is $xy = 4$.

Solution:

1. x-axis symmetry (replace y with -y): $x(-y) = 4 \implies -xy = 4$. Different. Not symmetric.

2. y-axis symmetry (replace x with -x): $(-x)y = 4 \implies -xy = 4$. Different. Not symmetric.

3. Origin symmetry (replace x with -x, y with -y): $(-x)(-y) = 4 \implies xy = 4$. Unchanged. Symmetric about the origin.

4. Line $y=x$ symmetry (interchange x and y): $yx = 4 \implies xy = 4$. Unchanged. Symmetric about the line $y=x$.



Parabola

The parabola is a U-shaped conic section with a unique and precise geometric definition. It is formed when a plane slices through a cone at an angle that is exactly parallel to the cone's generator line ($\alpha = \beta$).

Parabolas are found in many real-world applications, from the path of a projectile under gravity to the design of satellite dishes, car headlights, and suspension bridges, due to their remarkable reflective and structural properties.


Definition of a Parabola

A parabola is the set (or locus) of all points in a plane that are equidistant from a fixed point and a fixed line. The fixed point is called the focus, and the fixed line is called the directrix.

Let S be the focus and let line $l$ be the directrix. For any point P on the parabola, the defining property is:

Distance(P, S) = Perpendicular Distance(P, $l$)

PS = PM (where M is the foot of the perpendicular from P to the directrix)

Key Terminology

The definition of a parabola gives rise to several important features:

A diagram of a parabola showing its key features: the focus (a point), the directrix (a line), the axis of symmetry (a line), the vertex (a point), and the latus rectum (a line segment).

Derivation of the Standard Equation ($y^2 = 4ax$)

To derive the simplest equation for a parabola, we place it on the Cartesian plane in a standard position: let the vertex be at the origin $(0, 0)$ and the axis of symmetry be the x-axis. This means the parabola will open to the right or left.

Let's derive the equation for a parabola opening to the right.

Derivation setup for a parabola, showing a point P(x,y), the focus S(a,0), and the directrix x=-a. The distance PS is shown to be equal to the perpendicular distance PM.

Now, let $P(x, y)$ be any arbitrary point on the parabola. According to the definition of a parabola:

Distance(P, S) = Perpendicular Distance(P, $l$)

The distance PS can be found using the distance formula: $PS = \sqrt{(x-a)^2 + (y-0)^2}$.

The perpendicular distance from a point $(x, y)$ to the vertical line $x+a=0$ is simply $|x+a|$. Since the parabola opens right, $x$ will be non-negative, so this is $x+a$.

Setting these distances equal:

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

To remove the square root, we square both sides:

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

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

Cancel the common terms ($x^2$ and $a^2$) from both sides:

$-2ax + y^2 = 2ax$

Isolate $y^2$:

$y^2 = 4ax$

This is the standard equation for a parabola with its vertex at the origin, opening to the right.


Standard Equations of a Parabola

The equation of a parabola is simplest when its vertex is at the origin $(0, 0)$ and its axis of symmetry coincides with either the x-axis or the y-axis. There are four such standard forms. The constant $a$ (where $a > 0$) always represents the distance from the vertex to the focus.

1. Parabola Opening to the Right ($y^2 = 4ax$)

This is the standard form derived above.

Graph of the parabola y^2=4ax opening to the right, with vertex at the origin, focus at (a,0), and directrix at x=-a.

2. Parabola Opening to the Left ($y^2 = -4ax$)

This parabola opens towards the negative x-axis. The derivation is similar, but the focus is at $(-a, 0)$ and the directrix is $x=a$.

Graph of the parabola y^2=-4ax opening to the left, with vertex at the origin, focus at (-a,0), and directrix at x=a.

3. Parabola Opening Upwards ($x^2 = 4ay$)

Here, the roles of x and y are interchanged. The axis of symmetry is the y-axis, and the parabola opens towards the positive y-axis.

Graph of the parabola x^2=4ay opening upwards, with vertex at the origin, focus at (0,a), and directrix at y=-a.

4. Parabola Opening Downwards ($x^2 = -4ay$)

This parabola opens towards the negative y-axis.

Graph of the parabola x^2=-4ay opening downwards, with vertex at the origin, focus at (0,-a), and directrix at y=a.

Form $y^2 = 4ax$ $y^2 = -4ax$ $x^2 = 4ay$ $x^2 = -4ay$
Opens Right Left Upwards Downwards
Vertex (0, 0) (0, 0) (0, 0) (0, 0)
Focus (a, 0) (-a, 0) (0, a) (0, -a)
Directrix $x = -a$ $x = a$ $y = -a$ $y = a$
Axis of Symmetry x-axis ($y=0$) x-axis ($y=0$) y-axis ($x=0$) y-axis ($x=0$)
Length of Latus Rectum $4a$ $4a$ $4a$ $4a$
Equation of Latus Rectum $x = a$ $x = -a$ $y = a$ $y = -a$

Example 1. Find the coordinates of the focus, the equation of the directrix, and the length of the latus rectum for the parabola $y^2 = 12x$.

Answer:

Given: The equation of the parabola is $y^2 = 12x$.

Solution:

Step 1: The equation matches the form $y^2 = 4ax$, which opens right.

Step 2: By comparing, we have $4a = 12$, which gives $a = 3$.

Step 3: Using $a=3$, we find the properties:

  • Focus: $(a, 0) \implies$ (3, 0).
  • Directrix: $x = -a \implies$ $x = -3$.
  • Length of Latus Rectum: $4a = 4(3) = 12$.

Example 2. Find the equation of the parabola with its vertex at the origin and focus at (0, -4).

Answer:

Given: Vertex: $(0, 0)$ and Focus: $(0, -4)$.

Solution:

Step 1: The focus is on the negative y-axis, so the parabola opens downwards. The standard form is $x^2 = -4ay$.

Step 2: The distance from the vertex (0,0) to the focus (0,-4) is $a = 4$.

Step 3: Substitute $a=4$ into the equation: $x^2 = -4(4)y$.

The required equation is $x^2 = -16y$.


Example 3. Find the equation of the parabola with vertex at the origin, axis of symmetry along the x-axis, and passing through the point (2, 3).

Answer:

Given: Vertex: (0,0), Axis: x-axis, Point on parabola: (2,3).

Solution:

Step 1: Since the axis of symmetry is the x-axis and the vertex is at the origin, the equation is either $y^2 = 4ax$ or $y^2 = -4ax$.

Step 2: The point (2,3) has a positive x-coordinate. Since the point lies on the parabola, the parabola must open to the right. Thus, the correct form is $y^2 = 4ax$.

Step 3: The point (2,3) must satisfy the equation. Substitute $x=2$ and $y=3$ to find $a$.

$(3)^2 = 4a(2) \implies 9 = 8a \implies a = \frac{9}{8}$.

Step 4: Substitute the value of $a$ back into the standard form.

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

The required equation is $2y^2 = 9x$.



Ellipse

The ellipse is a conic section that results in a closed, oval-shaped curve. It is formed when a plane intersects a cone at an angle that is less steep than the cone's generator but not perpendicular to the cone's axis ($\beta < \alpha < 90^\circ$).

The ellipse has a precise geometric definition based on the sum of distances from two fixed points, which leads to its unique properties and applications in fields like astronomy (planetary orbits), architecture (whispering galleries), and engineering.


Definition of an Ellipse

An ellipse is the set (or locus) of all points in a plane for which the sum of the distances from two fixed points is a constant. The two fixed points are called the foci (plural of focus).

Let the two foci be $S_1$ and $S_2$. For any point P on the ellipse, the defining condition is:

Distance($P, S_1$) + Distance($P, S_2$) = Constant

This constant sum is always greater than the distance between the two foci and is equal to the length of the major axis ($2a$).

An ellipse is shown with its two foci, S1 and S2. A point P on the ellipse has distances PS1 and PS2, and the sum of these distances is constant.

Key Terminology and Concepts

A horizontal ellipse showing all key features: Foci (S1, S2), Center, Major Axis, Minor Axis, Vertices, and the relationship between a, b, and c.

Special Case (Circle): If the two foci $S_1$ and $S_2$ coincide at the center, then $c=0$. From the relation $a^2 = b^2 + c^2$, we get $a^2 = b^2$, so $a=b$. The major and minor axes are equal, and the ellipse becomes a circle. The eccentricity $e = c/a = 0/a = 0$. A circle is an ellipse with an eccentricity of 0.


Derivation of the Standard Equation

Let's derive the equation for an ellipse with its center at the origin $(0, 0)$ and its foci on the x-axis at $S_1(-c, 0)$ and $S_2(c, 0)$. Let the constant sum of distances be $2a$.

Let $P(x, y)$ be any point on the ellipse. By definition: $PS_1 + PS_2 = 2a$.

Using the distance formula:

$\sqrt{(x - (-c))^2 + (y-0)^2} + \sqrt{(x-c)^2 + (y-0)^2} = 2a$

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

Squaring both sides:

$(x+c)^2 + y^2 = 4a^2 - 4a\sqrt{(x-c)^2 + y^2} + ((x-c)^2 + y^2)$

$x^2+2cx+c^2+y^2 = 4a^2 - 4a\sqrt{(x-c)^2+y^2} + x^2-2cx+c^2+y^2$

Canceling terms and rearranging: $4cx - 4a^2 = -4a\sqrt{(x-c)^2 + y^2}$

Divide by 4: $cx - a^2 = -a\sqrt{(x-c)^2 + y^2}$

Square both sides again: $(cx - a^2)^2 = (-a)^2((x-c)^2 + y^2)$

$c^2x^2 - 2a^2cx + a^4 = a^2(x^2 - 2cx + c^2 + y^2)$

$c^2x^2 - 2a^2cx + a^4 = a^2x^2 - 2a^2cx + a^2c^2 + a^2y^2$

Cancel the $-2a^2cx$ term and rearrange: $a^4 - a^2c^2 = a^2x^2 - c^2x^2 + a^2y^2$

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

From the relationship $a^2 = b^2 + c^2$, we know that $b^2 = a^2 - c^2$. Substituting this in:

$a^2b^2 = x^2b^2 + a^2y^2$

Finally, divide the entire equation by $a^2b^2$:

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


Standard Equations of an Ellipse

The simplest equations for an ellipse occur when its center is at the origin $(0, 0)$ and its major and minor axes are aligned with the coordinate axes. The value $a$ is always the semi-major axis, and thus $a^2$ is always the larger denominator.

1. Major Axis along the x-axis (Horizontal Ellipse)

This ellipse is wider than it is tall. The larger denominator ($a^2$) is under the $x^2$ term.

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

... (i)

A horizontal ellipse with center at the origin. The vertices are at (+-a, 0) on the x-axis, and the foci are at (+-c, 0).

2. Major Axis along the y-axis (Vertical Ellipse)

This ellipse is taller than it is wide. The larger denominator ($a^2$) is under the $y^2$ term.

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

... (ii)

A vertical ellipse with center at the origin. The vertices are at (0, +-a) on the y-axis, and the foci are at (0, +-c).

Example 1. For the ellipse $\frac{x^2}{25} + \frac{y^2}{9} = 1$, find the lengths of the major and minor axes, the coordinates of the foci and vertices, the eccentricity, and the length of the latus rectum.

Answer:

Given: The equation is $\frac{x^2}{25} + \frac{y^2}{9} = 1$.

Solution:

Step 1: The larger denominator (25) is under $x^2$, so it is a horizontal ellipse. $a^2 = 25 \implies a = 5$ and $b^2 = 9 \implies b = 3$.

Step 2: Length of Major Axis = $2a = 10$. Length of Minor Axis = $2b = 6$.

Step 3: Find c: $c^2 = a^2 - b^2 = 25 - 9 = 16 \implies c = 4$.

Step 4: Coordinates:

  • Vertices: $(\pm a, 0) \implies (\pm 5, 0)$.
  • Foci: $(\pm c, 0) \implies (\pm 4, 0)$.

Step 5: Eccentricity: $e = \frac{c}{a} = \frac{4}{5} = 0.8$.

Step 6: Length of Latus Rectum: $\frac{2b^2}{a} = \frac{2(9)}{5} = \frac{18}{5} = 3.6$.


Example 2. Find the equation of the ellipse with vertices at $(\pm 6, 0)$ and foci at $(\pm 4, 0)$.

Answer:

Given: Vertices: $(\pm 6, 0)$ and Foci: $(\pm 4, 0)$.

Solution:

Step 1: Vertices and foci are on the x-axis, so it's a horizontal ellipse with center (0,0). The equation is $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$.

Step 2: From vertices, the distance from center to vertex is $a=6$. From foci, the distance from center to focus is $c=4$.

Step 3: Find $b^2$: $b^2 = a^2 - c^2 = 6^2 - 4^2 = 36 - 16 = 20$.

Step 4: Substitute $a^2=36$ and $b^2=20$ into the equation: $\frac{x^2}{36} + \frac{y^2}{20} = 1$.


Example 3. Find the equation of the ellipse with foci at $(0, \pm 3)$ and eccentricity $e = \frac{3}{5}$.

Answer:

Given: Foci: $(0, \pm 3)$ and Eccentricity $e = \frac{3}{5}$.

Solution:

Step 1: Foci are on the y-axis, so it's a vertical ellipse with center (0,0). The equation is $\frac{x^2}{b^2} + \frac{y^2}{a^2} = 1$.

Step 2: From foci, we know $c = 3$.

Step 3: Use the eccentricity formula $e = \frac{c}{a}$ to find $a$.

$\frac{3}{5} = \frac{3}{a} \implies 3a = 15 \implies a = 5$.

Step 4: Find $b^2$: $b^2 = a^2 - c^2 = 5^2 - 3^2 = 25 - 9 = 16$.

Step 5: Substitute $a^2=25$ and $b^2=16$ into the equation: $\frac{x^2}{16} + \frac{y^2}{25} = 1$.



Hyperbola

The hyperbola is the final of the four non-degenerate conic sections. It is formed when a plane slices through both nappes of a double-napped cone ($\alpha < \beta$), resulting in an open curve with two separate, unbounded branches that are mirror images of each other.

The geometric definition of a hyperbola is based on the *difference* of distances from two fixed points, which contrasts sharply with the ellipse's definition based on the *sum* of distances. This unique property makes hyperbolas essential in various scientific and technological fields.

Applications include navigation (the LORAN system was based on this principle), astronomy (describing the paths of some comets that are flung out of the solar system), and physics (representing shock waves or the paths of certain subatomic particles).


Definition of a Hyperbola

A hyperbola is the set (or locus) of all points in a plane for which the absolute difference of the distances from two fixed points remains constant. The two fixed points are called the foci.

Let the two fixed points be the foci $S_1$ and $S_2$. A point P is on the hyperbola if the result of subtracting one distance from the other is always the same positive number. The absolute value covers both branches of the hyperbola elegantly:

|Distance($P, S_1$) - Distance($P, S_2$)| = Constant ($2a$)

This constant difference, $2a$, must always be less than the distance between the foci, $2c$. This condition, $a < c$, ensures that the locus forms a two-branched hyperbola.

A hyperbola with two branches and two foci, S1 and S2. A point P on one branch has distances PS1 and PS2, and the absolute difference of these distances is constant.

Key Terminology and Concepts

A horizontal hyperbola showing all key features: Foci, Center, Transverse Axis, Conjugate Axis, Vertices, Asymptotes, and the central rectangle.

Derivation of the Standard Equation

Let's derive the equation for a hyperbola with its center at the origin $(0, 0)$ and its foci on the x-axis at $S_1(-c, 0)$ and $S_2(c, 0)$. The constant difference is $2a$.

Let $P(x, y)$ be any point on the hyperbola. By definition: $|PS_1 - PS_2| = 2a$.

The derivation proceeds almost identically to that of the ellipse, starting with:

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

After isolating one radical, squaring, simplifying, and squaring again (as done for the ellipse), we arrive at the step:

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

For a hyperbola, we have the relationship $c^2 = a^2 + b^2$, which means $b^2 = c^2 - a^2$. Substituting this in:

$x^2b^2 - a^2y^2 = a^2b^2$

Finally, divide the entire equation by $a^2b^2$ to get the standard form:

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


Standard Equations of a Hyperbola

The simplest equations for a hyperbola are obtained when its center is at the origin $(0, 0)$. The key to identifying the orientation is to look for the term with the positive coefficient. The variable in that term indicates the direction of the transverse axis.

1. Transverse Axis along the x-axis (Horizontal Hyperbola)

The branches open to the left and right. The $x^2$ term is positive.

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

... (i)

A horizontal hyperbola with its center at the origin. The vertices are at (+-a, 0) on the x-axis, and the foci are at (+-c, 0). The asymptotes are shown passing through the origin.

2. Transverse Axis along the y-axis (Vertical Hyperbola)

The branches open upwards and downwards. The $y^2$ term is positive.

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

... (ii)

A vertical hyperbola with its center at the origin. The vertices are at (0, +-a) on the y-axis, and the foci are at (0, +-c). The asymptotes are shown passing through the origin.

Example 1. For the hyperbola $\frac{x^2}{16} - \frac{y^2}{9} = 1$, find the coordinates of the foci and vertices, the eccentricity, and the equations of the asymptotes.

Answer:

Given: Equation is $\frac{x^2}{16} - \frac{y^2}{9} = 1$.

Solution:

Step 1: The $x^2$ term is positive, so it's a horizontal hyperbola. $a^2 = 16 \implies a = 4$ and $b^2 = 9 \implies b = 3$.

Step 2: Find c: $c^2 = a^2 + b^2 = 16 + 9 = 25 \implies c = 5$.

Step 3: Determine properties:

  • Vertices: $(\pm a, 0) \implies$ $(\pm 4, 0)$.
  • Foci: $(\pm c, 0) \implies$ $(\pm 5, 0)$.
  • Eccentricity: $e = \frac{c}{a} = \frac{5}{4}$.
  • Asymptotes: $y = \pm \frac{b}{a}x \implies$ $y = \pm \frac{3}{4}x$.

Example 2. Find the equation of the hyperbola with vertices at $(0, \pm 5)$ and foci at $(0, \pm 8)$.

Answer:

Given: Vertices: $(0, \pm 5)$ and Foci: $(0, \pm 8)$.

Solution:

Step 1: Vertices and foci are on the y-axis, so it's a vertical hyperbola with center (0,0). The equation is $\frac{y^2}{a^2} - \frac{x^2}{b^2} = 1$.

Step 2: From vertices, $a=5$. From foci, $c=8$.

Step 3: Find $b^2$: $b^2 = c^2 - a^2 = 8^2 - 5^2 = 64 - 25 = 39$.

Step 4: Substitute $a^2=25$ and $b^2=39$ into the equation: $\frac{y^2}{25} - \frac{x^2}{39} = 1$.


Example 3. Find the equation of the hyperbola with foci $(\pm 5, 0)$ and the length of the transverse axis is 8.

Answer:

Given: Foci: $(\pm 5, 0)$ and Length of Transverse Axis = 8.

Solution:

Step 1: Foci are on the x-axis, so it's a horizontal hyperbola with center (0,0). The equation is $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$.

Step 2: From foci, we know $c=5$.

Step 3: The length of the transverse axis is $2a$. So, $2a = 8 \implies a = 4$.

Step 4: Find $b^2$: $b^2 = c^2 - a^2 = 5^2 - 4^2 = 25 - 16 = 9$.

Step 5: Substitute $a^2=16$ and $b^2=9$ into the equation: $\frac{x^2}{16} - \frac{y^2}{9} = 1$.



Parametric Equations of Conics

In our study of curves, we have primarily used Cartesian equations, which establish a direct relationship between the $x$ and $y$ coordinates of points on the curve (e.g., $x^2 + y^2 = r^2$). An alternative and often powerful way to describe a curve is by using parametric equations.

This method defines the path of a point by describing its $x$ and $y$ coordinates as separate functions of a third, independent variable called a parameter. This approach is particularly useful in physics to describe motion, where the parameter is often time, $t$. As time progresses, the parametric equations tell you the exact $(x, y)$ position of a moving object.


Parametric Representation of a Curve

A curve can be represented by a pair of equations where the $x$ and $y$ coordinates are given as functions of a parameter, let's say $t$:

$x = f(t)$ and $y = g(t)$

For each value of the parameter $t$ within a specified domain, we get a unique point $(x, y)$ on the curve. The entire curve is traced out as $t$ varies through its range of values. The pair of equations is called a parametric representation of the curve.

Converting from Parametric to Cartesian Form

To convert from a parametric representation back to the more familiar Cartesian form, the goal is to eliminate the parameter. This creates a single equation that directly relates $x$ and $y$. The method for elimination depends on the form of the equations:


Standard Parametric Equations for Conics

Each of the standard conic sections has a conventional parametric representation that is particularly useful and arises naturally from its geometry.

1. Circle: $x^2 + y^2 = r^2$

For a circle centered at the origin, the most natural parameter is the angle $\theta$ that a line segment from the origin to a point on the circle makes with the positive x-axis. This is the standard angle in polar coordinates.

A circle centered at the origin with radius r. A point P(x,y) is on the circle, and the angle theta is shown. A right triangle demonstrates that x = r cos(theta) and y = r sin(theta).

Using basic trigonometry in the right-angled triangle shown in the diagram, the coordinates $(x, y)$ can be expressed as:

$x = r \cos \theta$

$y = r \sin \theta$

... (i)

The entire circle is traced once as the parameter $\theta$ varies from $0$ to $2\pi$. We can verify this by eliminating the parameter using the identity $\cos^2 \theta + \sin^2 \theta = 1$.

From the equations, $\cos \theta = \frac{x}{r}$ and $\sin \theta = \frac{y}{r}$. Substituting into the identity gives $(\frac{x}{r})^2 + (\frac{y}{r})^2 = 1 \implies \frac{x^2}{r^2} + \frac{y^2}{r^2} = 1 \implies x^2 + y^2 = r^2$.

2. Parabola: $y^2 = 4ax$

For a parabola, a simple algebraic parameter $t$ is used. The most common parametrization is derived to automatically satisfy the Cartesian equation.

$x = at^2$

$y = 2at$

... (ii)

Here, the parameter $t$ can be any real number ($-\infty < t < \infty$). Any point of the form $(at^2, 2at)$ is often referred to as 'the point $t$' on the parabola. We can verify this by substituting these expressions into $y^2 = 4ax$:

LHS: $y^2 = (2at)^2 = 4a^2t^2$.

RHS: $4ax = 4a(at^2) = 4a^2t^2$.

Since LHS = RHS, the parametrization is valid.

3. Ellipse: $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$

Similar to the circle, the ellipse uses an angular parameter $\theta$, often called the eccentric angle. The representation is a scaled version of the circle's parametrization, stretched horizontally by a factor of 'a' and vertically by a factor of 'b'.

$x = a \cos \theta$

$y = b \sin \theta$

... (iii)

The entire ellipse is traced as $\theta$ varies from $0$ to $2\pi$. We verify this using the same identity as the circle. From the equations, $\cos \theta = \frac{x}{a}$ and $\sin \theta = \frac{y}{b}$. Substituting into $\cos^2 \theta + \sin^2 \theta = 1$ gives $(\frac{x}{a})^2 + (\frac{y}{b})^2 = 1$, which is the Cartesian equation.

4. Hyperbola: $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$

The hyperbola's parametrization is based on the Pythagorean identity involving secant and tangent: $\sec^2 \theta - \tan^2 \theta = 1$.

$x = a \sec \theta$

$y = b \tan \theta$

... (iv)

Here, the parameter $\theta$ varies over a range like $0 \le \theta < 2\pi$, excluding the values where the functions are undefined (e.g., $\theta = \frac{\pi}{2}, \frac{3\pi}{2}$). To verify, we have $\sec \theta = \frac{x}{a}$ and $\tan \theta = \frac{y}{b}$. Substituting into the identity gives $(\frac{x}{a})^2 - (\frac{y}{b})^2 = 1$.


Example 1. Find the Cartesian equation of the curve represented by the parametric equations $x = 3t^2$ and $y = 6t$.

Answer:

Given: $x = 3t^2$ and $y = 6t$.

Solution:

Step 1: Solve the linear equation for $t$. From $y = 6t$, we get $t = \frac{y}{6}$.

Step 2: Substitute this expression for $t$ into the other equation, $x = 3t^2$.

$x = 3\left(\frac{y}{6}\right)^2 = 3\left(\frac{y^2}{36}\right) = \frac{y^2}{12}$.

Step 3: Rearrange the equation: $12x = y^2$.

The Cartesian equation is $y^2 = 12x$, which is a parabola.


Example 2. Find the Cartesian equation of the curve represented by $x = 5 \cos \theta$ and $y = 2 \sin \theta$.

Answer:

Given: $x = 5 \cos \theta$ and $y = 2 \sin \theta$.

Solution:

Step 1: Isolate the trigonometric functions.

$\cos \theta = \frac{x}{5}$

$\sin \theta = \frac{y}{2}$

Step 2: Use the identity $\cos^2 \theta + \sin^2 \theta = 1$.

Substitute the expressions from Step 1 into the identity:

$\left(\frac{x}{5}\right)^2 + \left(\frac{y}{2}\right)^2 = 1$

Step 3: Simplify the equation.

$\frac{x^2}{25} + \frac{y^2}{4} = 1$

The Cartesian equation is $\frac{x^2}{25} + \frac{y^2}{4} = 1$, which is an ellipse.



Applications

The study of conic sections—parabolas, ellipses, and hyperbolas—is far more than a theoretical exercise in mathematics. These curves appear all around us in the natural world and are fundamental to the design of countless technologies. Their unique geometric and reflective properties are harnessed in fields ranging from astronomy and physics to engineering and medicine.


Applications of the Parabola

The most important property of the parabola is its ability to focus parallel rays to a single point, or conversely, to turn rays from a single point into a parallel beam. This is known as its reflective property.

A diagram illustrating the reflective property of a parabola. Rays approaching parallel to the axis of symmetry are all reflected to the focus point. Conversely, rays originating from the focus are reflected into a parallel beam.

This single property leads to numerous applications:

Example 1. A satellite dish is in the shape of a parabolic reflector. The dish is 12 feet across at its opening and 2 feet deep. How far from the vertex (the bottom of the dish) should the receiver be placed?

Answer:

Problem Analysis:

The receiver should be placed at the focus of the parabola. We need to find the coordinates of the focus.

Solution:

Step 1: Set up the coordinate system.

Let's place the vertex of the parabola at the origin $(0, 0)$. Since the dish opens upwards, its equation will be of the standard form $x^2 = 4ay$.

Step 2: Find a point on the parabola.

The dish is 12 feet across, which means the radius at the opening is half of that, or 6 feet. The depth is 2 feet. This gives us a point on the rim of the dish at $(6, 2)$.

A cross-section of the satellite dish shown as a parabola with vertex at (0,0) and a point on the rim at (6,2).

Step 3: Solve for 'a'.

Since the point $(6, 2)$ lies on the parabola, it must satisfy the equation $x^2 = 4ay$.

Substitute $x=6$ and $y=2$ into the equation:

$(6)^2 = 4a(2)$

$36 = 8a$

$a = \frac{36}{8} = \frac{9}{2} = 4.5$

Step 4: Locate the focus.

For a parabola of the form $x^2 = 4ay$, the focus is located at the point $(0, a)$.

Therefore, the focus is at $(0, 4.5)$.

This means the receiver should be placed 4.5 feet from the vertex along the axis of symmetry.

The receiver should be placed 4.5 feet from the bottom of the dish.


Applications of the Ellipse

The ellipse is defined by the property that the sum of the distances from any point on the curve to two fixed foci is constant. This leads to its own unique reflective property and other important applications.

Example 2. A "whispering gallery" is to be constructed in the shape of a semi-elliptical arch. The room is 100 feet long and 40 feet high at the center. How far from the center along the floor should two people stand to be at the foci of the ellipse?

Answer:

Problem Analysis:

We need to find the locations of the foci of the ellipse that defines the room's shape.

Solution:

Step 1: Set up the coordinate system.

Place the center of the ellipse at the origin $(0, 0)$. The length of the room corresponds to the major axis, and the height corresponds to the semi-minor axis.

The equation will be of the form $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$.

Step 2: Find 'a' and 'b'.

The total length is 100 feet, which is the length of the major axis ($2a$).

$2a = 100 \implies a = 50$ feet.

The height at the center is 40 feet, which is the length of the semi-minor axis ($b$).

$b = 40$ feet.

Step 3: Find 'c' to locate the foci.

Use the relationship for an ellipse: $c^2 = a^2 - b^2$.

$c^2 = (50)^2 - (40)^2 = 2500 - 1600 = 900$

$c = \sqrt{900} = 30$ feet.

Step 4: Determine the location of the foci.

For a horizontal ellipse centered at the origin, the foci are located at $(\pm c, 0)$.

This means the foci are 30 feet to the left and 30 feet to the right of the center.

The two people should stand 30 feet from the center along the major axis (the floor).


Applications of the Hyperbola

The hyperbola, defined by the constant *difference* of distances to two foci, has applications that often involve locating positions or describing certain physical phenomena.

Example 3. Two listening stations are located at points A(-8, 0) and B(8, 0) (units in km). They detect an explosion, and station B receives the sound 10 seconds after station A. Assuming sound travels at 0.34 km per second, determine the equation of the hyperbola on which the explosion must have occurred.

Answer:

Problem Analysis:

The listening stations are the foci of the hyperbola. The time difference gives us the constant distance difference, $2a$.

Solution:

Step 1: Determine the constants 'c' and 'a'.

The foci are at $(\pm 8, 0)$, so the center is at the origin and the distance from the center to a focus is $c = 8$ km.

The constant difference in distance is the difference in time multiplied by the speed of sound.

$2a = (\text{time difference}) \times (\text{speed}) = 10 \text{ s} \times 0.34 \text{ km/s} = 3.4$ km.

$a = \frac{3.4}{2} = 1.7$ km.

Step 2: Find 'b^2'.

Use the relationship for a hyperbola: $b^2 = c^2 - a^2$.

$b^2 = 8^2 - (1.7)^2 = 64 - 2.89 = 61.11$.

Step 3: Write the equation of the hyperbola.

Since the foci are on the x-axis, the hyperbola is horizontal. The equation is of the form $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$.

Substituting our values for $a^2 = (1.7)^2 = 2.89$ and $b^2 = 61.11$:

$\frac{x^2}{2.89} - \frac{y^2}{61.11} = 1$

The equation of the hyperbola is $\frac{x^2}{2.89} - \frac{y^2}{61.11} = 1$.


Applications of the Circle

As a special case of the ellipse (with an eccentricity of zero), the circle is the most common and fundamental shape, with applications that are almost too numerous to list.

The study of conic sections provides a powerful mathematical language to describe and utilize these fundamental shapes that are woven into the fabric of our universe and technology.