JEE Main 2026 — Jan 21, Shift 1 — Mathematics solution

JEE Main2026Jan 21, Shift 1Mathematics

Let O be the vertex of the parabola $x^{2} = 4 y$ and Q be any point on it. Let the locus of the point P, which divides the line segment OQ internally in the ratio $2 : 3$ be the conic C. Then the equation of the chord of C, which is bisected at the point $(1, 2)$ , is:

A
$5 x - y - 3 = 0$
B
$4 x - 5 y + 6 = 0$
C
$x - 2 y + 3 = 0$
D
$5 x - 4 y + 3 = 0$

Solution

Parametrise the parabola as $Q = (2 t, t^{2})$ . The internal section formula (with O at origin, ratio $2 : 3$ ) gives $P = (\frac{4 t}{5}, \frac{2 t ^{2}}{5}) = (h, k)$ .

Eliminating $t$ (using $t = 5 h /4$ ): $k = \frac{2}{5} \cdot \frac{25 h ^{2}}{16}$ , i.e. $5 x^{2} = 8 y$ — the locus $C$ .

For a chord of $5 x^{2} = 8 y$ bisected at $(x_{1}, y_{1})$ , the equation is $T = S_{1}$ : $5 x x_{1} - 4 (y + y_{1}) = 5 x_{1}^{2} - 8 y_{1}$ .

At $(1, 2)$ : $5 x - 4 (y + 2) = 5 - 16 \Rightarrow 5 x - 4 y + 3 = 0$ .

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