JEE Main 2026 — Jan 21, Shift 2 — Mathematics solution

JEE Main2026Jan 21, Shift 2Mathematics

Let $y^{2} = 12 x$ be the parabola with its vertex at O. Let P be a point on the parabola and A be a point on the x-axis such that $∠ O P A = 90 d e g;$ . Then the locus of the centroid of such triangles OPA is :

A
$y^{2} - 6 x + 4 = 0$
B
$y^{2} - 9 x + 6 = 0$
C
$y^{2} - 2 x + 8 = 0$
D
$y^{2} - 4 x + 8 = 0$

Solution

Parametrise $P = (3 t^{2}, 6 t)$ on $y^{2} = 12 x$ (with $4 a = 12 \Rightarrow a = 3$ ). The slope of $O P$ is $\frac{6 t}{3 t ^{2}} = \frac{2}{t}$ , so for $O P ⊥ P A$ the slope of $P A$ is $- \frac{t}{2}$ .

Line $P A$ : $y - 6 t = - \frac{t}{2} (x - 3 t^{2})$ . Putting $y = 0$ (since A is on the x-axis) gives $A = (12 + 3 t^{2}, 0)$ .

Centroid $G (h, k)$ of $△ O P A$ : $3 h = 0 + 3 t^{2} + (12 + 3 t^{2}) = 6 t^{2} + 12$ , $3 k = 6 t$ , so $t = \frac{k}{2}$ and $h = 2 t^{2} + 4 = \frac{k ^{2}}{2} + 4$ .

Eliminate $t$ : $k^{2} = 2 h - 8$ , i.e. $y^{2} - 2 x + 8 = 0$ .

← Back to full paper

Practice more Mathematics

Concept-wise practice with instant solutions on Fundamenthol.

Start practicing →