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| Definition of Locus | Finding the Equation of a Locus satisfying Given Conditions | |
Locus and its Equation
Definition of Locus
The term locus (pronounced LOW-kus) comes from Latin, meaning "place" or "location." In geometry, a locus is not just any place, but a very specific one: it refers to the set of all points that satisfy a given geometric condition or a set of conditions. Crucially, the locus must include *all* points that satisfy the condition and *only* those points.
Think of a locus as the path traced by a point as it moves according to a specific rule. For example, if a point moves such that it is always a fixed distance from a stationary point, the path it traces is a circle. The circle, in this case, is the locus of the moving point.
Understanding the Definition
To fully grasp the concept of a locus, consider these aspects:
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Set of all points: The definition requires that the locus includes every single point that meets the given geometric condition(s). No valid point should be left out.
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Only those points: This is the restrictive part. The locus must contain *only* the points that satisfy the condition(s). Any point that does not meet the criteria must not be on the locus.
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Geometric condition(s): These are the rules, constraints, or relationships that the points on the locus must adhere to. These conditions relate the position of the point (or points) to other fixed points, lines, curves, or geometric figures.
The process of finding a locus involves identifying the points that fit the description and then typically expressing this set of points as a geometric shape or an algebraic equation.
Examples of Loci
Let's look at some common examples of loci, illustrating how a simple geometric condition defines a specific shape:
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Condition: The set of all points in a plane that are a fixed distance $r$ from a single fixed point C.
Locus: This condition describes points whose distance from C is always constant. By definition, this is a circle with center C and radius $r$.
If C is at the origin (0, 0) and the fixed distance is $r$, the locus is the set of points P$(x, y)$ such that $\sqrt{(x-0)^2 + (y-0)^2} = r$, which simplifies to $x^2 + y^2 = r^2$. This is the equation of the circle.
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Condition: The set of all points in a plane that are equidistant from two distinct fixed points A and B.
Locus: Any point P on this locus must satisfy the condition $PA = PB$. The set of all such points forms the perpendicular bisector of the line segment joining A and B.
If $A(x_1, y_1)$ and $B(x_2, y_2)$, and $P(x, y)$, then $PA^2 = PB^2$. Using the distance formula squared: $(x - x_1)^2 + (y - y_1)^2 = (x - x_2)^2 + (y - y_2)^2$. Expanding and simplifying this will give a linear equation in $x$ and $y$, representing a straight line - the perpendicular bisector.
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Condition: The set of all points in a plane that are equidistant from a fixed line (called the directrix) and a fixed point (called the focus) not lying on the line.
Locus: This condition defines a parabola. If L is the directrix and F is the focus, any point P on the parabola satisfies $\text{distance}(P, F) = \text{distance}(P, \text{line } L)$.
For example, the locus of points equidistant from the point $(a, 0)$ and the line $x = -a$ is the parabola $y^2 = 4ax$.
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Condition: The set of all points in a plane such that the sum of the distances from two distinct fixed points (called the foci) is a constant value.
Locus: This condition defines an ellipse. If $F_1$ and $F_2$ are the foci, any point P on the ellipse satisfies $PF_1 + PF_2 = \text{constant}$.
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Condition: The set of all points in a plane such that the absolute difference of the distances from two distinct fixed points (called the foci) is a constant value.
Locus: This condition defines a hyperbola. If $F_1$ and $F_2$ are the foci, any point P on the hyperbola satisfies $|PF_1 - PF_2| = \text{constant}$.
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Condition: The set of all points in a plane that are equidistant from two intersecting lines.
Locus: Any point P on this locus must satisfy $\text{distance}(P, \text{line } L_1) = \text{distance}(P, \text{line } L_2)$. The set of all such points forms the pair of angle bisectors of the angles formed by the two lines. These two bisectors are perpendicular to each other.
In coordinate geometry, the study of loci primarily involves finding the equation of a locus. This is an algebraic equation relating the coordinates $(x, y)$ of a general point P on the locus. The equation must be satisfied by the coordinates of every point that lies on the locus and must NOT be satisfied by the coordinates of any point that does not lie on the locus. Finding the equation of a locus is a key technique for analyzing geometric problems algebraically.
Locus Summary (For Competitive Exams)
Definition:
The set of all points (and only those points) satisfying a given geometric condition(s).
Key Concept:
The path traced by a point moving under specific constraints.
Equation of Locus:
An algebraic equation in terms of coordinates ($x, y$) that represents all points on the locus and no others.
Examples:
- Equidistant from a point $\implies$ Circle
- Equidistant from two points $\implies$ Perpendicular Bisector
- Equidistant from a point and a line $\implies$ Parabola
- Sum of distances from two points is constant $\implies$ Ellipse
- Difference of distances from two points is constant $\implies$ Hyperbola
- Equidistant from two intersecting lines $\implies$ Angle Bisectors
Finding the Equation of a Locus:
Let the general point on the locus be P$(x, y)$. Use the given geometric condition(s) to write an equation involving the coordinates $(x, y)$ and the coordinates/parameters of any fixed entities (points, lines, etc.) involved in the condition. Simplify this equation algebraically to obtain the equation of the locus.
Finding the Equation of a Locus satisfying Given Conditions
The most significant aspect of working with loci in coordinate geometry is translating a given geometric description of a point's movement or position into an algebraic equation. This equation, expressed in terms of the coordinates $(x, y)$, uniquely represents the set of all points that satisfy the specified condition(s) and is called the equation of the locus.
The process of finding the equation involves a systematic approach that converts geometric relationships into algebraic ones using coordinate geometry formulas.
Steps to Find the Equation of a Locus
To find the equation of the locus of a point satisfying certain geometric conditions, follow these general steps:
| Step | Action | Details |
|---|---|---|
| 1 | Assume a General Point | Assume that $P(h, k)$ is any arbitrary point on the locus whose equation is to be found. Some prefer to use $P(x, y)$ directly from the start, but using $(h, k)$ temporarily can help distinguish the coordinates of the moving point from any fixed coordinates $(x_1, y_1)$, $(x_2, y_2)$, etc., or general variables like $x$ and $y$ that might appear in line equations. |
| 2 | Write the Geometric Condition(s) | Clearly state or write down the geometric condition(s) that the point $P(h, k)$ must satisfy as given in the problem statement. It is often helpful to draw a rough diagram to visualize the given points/lines and the moving point $P$. |
| 3 | Translate to Algebraic Equation(s) | Convert the geometric condition(s) into an algebraic equation or equations involving $h$, $k$, and the coordinates/parameters of any fixed points, lines, or other geometric figures given in the problem. Use appropriate formulas from coordinate geometry, such as:
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| 4 | Simplify the Algebraic Equation | Simplify the equation(s) obtained in Step 3. This often involves squaring both sides to remove square roots (be careful about introducing extraneous solutions if the geometric condition itself wasn't about squared distances), expanding binomials, combining like terms, and rearranging the equation into a standard form (e.g., $Ax+By+C=0$ for a line, $x^2+y^2+Dx+Ey+F=0$ for a circle, etc.). The goal is to obtain a simplified equation in $h$ and $k$. |
| 5 | Generalize to (x, y) | Replace the temporary variables $h$ with $x$ and $k$ with $y$ in the final simplified equation. The resulting equation in terms of $x$ and $y$ is the equation of the required locus. This equation holds true for every point $(x, y)$ that satisfies the original geometric condition. |
Example 1. Find the equation of the locus of a point which moves such that it is always equidistant from the points A(2, 3) and B(4, 5).
Answer:
Step 1: Assume a General Point
Let $P(h, k)$ be any point on the locus.
Step 2: Write the Geometric Condition
The condition given is that the distance from $P$ to point A is equal to the distance from $P$ to point B.
PA = PB
(Given condition)
Step 3: Translate to Algebraic Equation
Using the distance formula, $d = \sqrt{(x_2-x_1)^2 + (y_2-y_1)^2}$, we express the distances PA and PB algebraically. It is often easier to work with the square of the distance to eliminate the square root immediately.
$PA^2 = (h - 2)^2 + (k - 3)^2$
$PB^2 = (h - 4)^2 + (k - 5)^2$
Since $PA = PB$, squaring both sides gives $PA^2 = PB^2$.
$(h - 2)^2 + (k - 3)^2 = (h - 4)^2 + (k - 5)^2$
($PA^2 = PB^2$)
Step 4: Simplify the Algebraic Equation
Expand the squared terms using $(a-b)^2 = a^2 - 2ab + b^2$:
$(h^2 - 4h + 4) + (k^2 - 6k + 9) = (h^2 - 8h + 16) + (k^2 - 10k + 25)$
... (i)
Combine constant terms on each side of equation (i):
$h^2 - 4h + k^2 - 6k + 13 = h^2 - 8h + k^2 - 10k + 41$
... (ii)
Subtract $h^2$ and $k^2$ from both sides of equation (ii):
$-4h - 6k + 13 = -8h - 10k + 41$
... (iii)
Move all terms to the left side of equation (iii) to set the equation to zero:
$(-4h + 8h) + (-6k + 10k) + (13 - 41) = 0$
$4h + 4k - 28 = 0$
... (iv)
Divide equation (iv) by 4 to simplify:
$h + k - 7 = 0$
... (v)
Equation (v) is the simplified relationship between $h$ and $k$ for any point $P(h, k)$ on the locus.
Step 5: Generalize to (x, y)
Replace $h$ with $x$ and $k$ with $y$ to get the equation of the locus in terms of standard coordinate variables:
$\mathbf{x + y - 7 = 0}$
... (vi)
This is the equation of a straight line, which is indeed the perpendicular bisector of the segment joining A(2, 3) and B(4, 5).
Example 2. Find the equation of the locus of a point $P$ such that the distance from $P$ to the point A(3, 0) is always twice the distance from $P$ to the point B(-3, 0).
Answer:
Step 1: Assume a General Point
Let $P(h, k)$ be any point on the locus.
Step 2: Write the Geometric Condition
The condition is that the distance PA is twice the distance PB.
PA = 2 $\times$ PB
(Given condition)
Step 3: Translate to Algebraic Equation
To eliminate the square roots from the distance formula, square both sides:
$PA^2 = (2 \times PB)^2 = 4 \times PB^2$
... (i)
Using the distance formula squared:
$PA^2 = (h - 3)^2 + (k - 0)^2 = (h - 3)^2 + k^2$
... (ii)
$PB^2 = (h - (-3))^2 + (k - 0)^2 = (h + 3)^2 + k^2$
... (iii)
Substitute expressions for $PA^2$ and $PB^2$ from (ii) and (iii) into equation (i):
$(h - 3)^2 + k^2 = 4 [(h + 3)^2 + k^2]$
... (iv)
Step 4: Simplify the Algebraic Equation
Expand the squared terms in equation (iv):
$(h^2 - 6h + 9) + k^2 = 4 [ (h^2 + 6h + 9) + k^2 ]$
... (v)
Distribute the 4 on the right side of equation (v):
$h^2 - 6h + 9 + k^2 = 4h^2 + 24h + 36 + 4k^2$
... (vi)
Move all terms to the right side of equation (vi) to set it to zero:
$0 = (4h^2 - h^2) + (24h + 6h) + (4k^2 - k^2) + (36 - 9)$
$0 = 3h^2 + 30h + 3k^2 + 27$
... (vii)
Divide equation (vii) by 3 to simplify:
$0 = h^2 + 10h + k^2 + 9$
... (viii)
Rearrange equation (viii) into a standard form:
$h^2 + k^2 + 10h + 9 = 0$
... (ix)
Equation (ix) is the simplified relationship between $h$ and $k$ for any point $P(h, k)$ on the locus.
Step 5: Generalize to (x, y)
Replace $h$ with $x$ and $k$ with $y$ to get the equation of the locus:
$\mathbf{x^2 + y^2 + 10x + 9 = 0}$
... (x)
This equation represents a circle (specifically, the Circle of Apollonius for points A and B and ratio 2:1). We can complete the square to find its center and radius: $(x^2 + 10x + 25) + y^2 + 9 - 25 = 0 \implies (x+5)^2 + y^2 = 16 = 4^2$. This is a circle with center $(-5, 0)$ and radius $4$.
Example 3. Find the equation of the locus of a point that moves such that its distance from the x-axis is equal to its distance from the point (2, 3).
Answer:
Step 1: Assume a General Point
Let $P(h, k)$ be any point on the locus.
Step 2: Write the Geometric Condition
The distance of a point $P(h, k)$ from the x-axis is the absolute value of its y-coordinate, i.e., $|k|$.
The distance of $P(h, k)$ from the fixed point $F(2, 3)$ is $\sqrt{(h-2)^2 + (k-3)^2}$ using the distance formula.
The given condition is: Distance from x-axis = Distance from F(2, 3).
$|k| = \sqrt{(h-2)^2 + (k-3)^2}$
(Given condition)
Step 3: Translate to Algebraic Equation
Square both sides to eliminate the square root:
$(|k|)^2 = \left(\sqrt{(h-2)^2 + (k-3)^2}\right)^2$
... (i)
Since $(|k|)^2 = k^2$, equation (i) becomes:
$k^2 = (h-2)^2 + (k-3)^2$
... (ii)
Step 4: Simplify the Algebraic Equation
Expand the squared terms in equation (ii):
$k^2 = (h^2 - 4h + 4) + (k^2 - 6k + 9)$
... (iii)
Subtract $k^2$ from both sides of equation (iii):
$0 = h^2 - 4h + 4 - 6k + 9$
... (iv)
Combine constant terms in equation (iv):
$0 = h^2 - 4h - 6k + 13$
... (v)
Rearrange equation (v) into a standard form:
$h^2 - 4h - 6k + 13 = 0$
... (vi)
Equation (vi) is the simplified relationship between $h$ and $k$ for any point $P(h, k)$ on the locus.
Step 5: Generalize to (x, y)
Replace $h$ with $x$ and $k$ with $y$ to get the equation of the locus:
$\mathbf{x^2 - 4x - 6y + 13 = 0}$
... (vii)
This equation represents a parabola. The fixed point (2, 3) is the focus, and the fixed line (x-axis, $y=0$) is the directrix. The equation is of the form $(x-a)^2 = 4b(y-c)$ or $(y-c)^2 = 4b(x-a)$, characteristic of a parabola.
Locus Equation Steps (For Competitive Exams)
Quick Guide:
- Assume the point P$(h, k)$ on the locus.
- Write the given geometric condition mathematically involving $P$ and any fixed entities.
- Translate the condition into an algebraic equation using coordinate formulas (distance, slope, etc.). Square to remove $\sqrt{}$ if necessary.
- Simplify the equation involving $h$ and $k$.
- Replace $h$ with $x$ and $k$ with $y$ to get the final equation of the locus.
Common Conditions & Formulas:
- Distance between two points: $\sqrt{(x_2-x_1)^2 + (y_2-y_1)^2}$
- Distance from point to line $Ax+By+C=0$: $\frac{|Ax_0+By_0+C|}{\sqrt{A^2+B^2}}$
- Slope $m = \frac{y_2-y_1}{x_2-x_1}$
- Perpendicular lines: $m_1m_2 = -1$
Resulting Loci (Examples):
- $x^2+y^2+Dx+Ey+F=0$ (Circle)
- $Ax+By+C=0$ (Line)
- $Ax^2+Bxy+Cy^2+Dx+Ey+F=0$ (Conic Section - Parabola, Ellipse, Hyperbola depending on coefficients)