VIT - VITEEE 2025
National level exam conducted by VIT University, Vellore | Ranked #11 by NIRF for Engg. | NAAC A++ Accredited | Last Date to Apply: 7th April | NO Further Extensions!
Locus and its Equation is considered one of the most asked concept.
176 Questions around this concept.
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
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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
National level exam conducted by VIT University, Vellore | Ranked #11 by NIRF for Engg. | NAAC A++ Accredited | Last Date to Apply: 7th April | NO Further Extensions!
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Tangents PA and PB are drawn from a point P to the ellipse. Find the locus of a point P, if the area of the triangle formed by the chord of contact AB and axes of co-ordinates are constant (c).
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
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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