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Locus - Practice Questions & MCQ

Edited By admin | Updated on Sep 18, 2023 18:34 AM | #JEE Main

Quick Facts

  • Locus and its Equation is considered one of the most asked concept.

  • 169 Questions around this concept.

Solve by difficulty

Let $O$ be the vertex and $Q$ be any point on the parabola, $x^2=8 y$. If the point $P$ divides the line segment $O Q$ internally in the ratio $1: 3$, then the locus of $P$ is :

The locus of the centres of the circles, which touch the circle,

$x^{2}+y^{2}=1$ externally , also touch the y-axis and lie in the

first quadrant, is :

A variable line drawn through the point $(1, 3)$ meets the x-axis at $A$ and the y-axis at $B$. If the rectangle $OAPB$ is completed, where ‘O’ is the origin, then the locus of ‘P’ is

The Cartesian equation of the curve $x=7+4 \cos \alpha, y=-3+4 \sin \alpha$ is

The locus of a point for which x = 0 is
 

The locus of a point for which y = 0, z = 0 is
 

If R be the point (0, 1) and Q is a variable point on $y=4x+2$. Then the locus of the mid-point of QR is

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The maximum possible value of $x^{2}+y^{2}-4x-6y,x,y$ real, subject to the condition $\left | x+y \right |+\left | x-y \right |=4$ is

Concepts Covered - 1

Locus and its Equation

Locus and its Equation

When a point moves in a plane under certain geometrical conditions, then the point traces out a path, This path of the moving point is known as the locus of this point.   

For example, let a point O(0,0) is a fixed point (i.e. origin) and a variable point P (x, y) is in the same plane. If point P moves in such a way that the distance OP is constant r, then point P traces out a circle whose center is O(0, 0) and radius is r.

  
 

Steps to Finding the Equation of Locus

Consider the point (h, k) whose locus is to be found.

Express the given condition as an equation in terms of the known quantities and unknown parameters.

Eliminate the parameters so that the resultant equation consists only locus coordinates h, k, and known quantities.

Now, replace the locus coordinate (h, k) with (x, y) in the resultant equation.

Illustriation

Find the equation of the locus of the point which is at a constant distance of 5 units from a point $(2,3)$
Solution
Let $A=(2,3)$ and $B=(h, k)$
$\mathrm{AB}=$ constant $=5$
$(A B)^2=5^2=25$
$(A B)^2=(h-2)^2+(k-3)^2=25$

$
h^2-4 h+4+k^2-6 k+9=25
$

The equation of locus is

$
x^2+y^2-4 x-6 y-12=0
$
In next chapter, we will see that this equation represents circle with centre at the point $(2,3)$ with radius 5 unit

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Locus and its Equation

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Books

Reference Books

Locus and its Equation

Mathematics for Joint Entrance Examination JEE (Advanced) : Coordinate Geometry

Page No. : 1.21

Line : 49

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