JEE Main 2026 — Jan 22, Shift 2 — Mathematics solution

JEE Main2026Jan 22, Shift 2Mathematics

Let the locus of the mid-point of the chord through the origin $O$ of the parabola $y^{2} = 4 x$ be the curve $S$ . Let $P$ be any point on $S$ . Then the locus of the point, which internally divides $O P$ in the ratio $3 : 1$ , is :

A
$3 y^{2} = 2 x$
B
$2 y^{2} = 3 x$
C
$3 x^{2} = 2 y$
D
$2 x^{2} = 3 y$

Solution

A point on $y^{2} = 4 x$ can be written as $P_{0} (t^{2}, 2 t)$ . The midpoint of $O P_{0}$ is $M (h, k) = (\frac{t ^{2}}{2}, t)$ , so $k^{2} = t^{2} = 2 h$ . The curve $S$ is $y^{2} = 2 x$ .

Now take a general point on $S$ as $P (t^{2} /2, t)$ and let $R (h, k)$ divide $O P$ internally in ratio $3 : 1$ :

$h = \frac{3 \cdot ( t ^{2} /2 )}{4} = \frac{3 t ^{2}}{8}$ , $k = \frac{3 t}{4}$ .

So $t = \frac{4 k}{3}$ and $t^{2} = \frac{8 h}{3}$ , giving $\frac{16 k ^{2}}{9} = \frac{8 h}{3} \Rightarrow 2 k^{2} = 3 h$ .

Locus: $2 y^{2} = 3 x$ .

← Back to full paper

Practice more Mathematics

Concept-wise practice with instant solutions on Fundamenthol.

Start practicing →