Aspire Faculty ID #17904 · Topic: JEE Main 2026 (21 January Morning Shift) · Just now
JEE Main 2026 (21 January Morning Shift)

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

Solution

$h = \frac{4t}{5}$



$k = \frac{2t^2}{5} = \frac{2}{5}\left(\frac{5h}{4}\right)^2$

$8k = 5h^2$

$\Rightarrow 5x^2 = 8y$

$T = S_1$

$5(xx_1) - 4(y + y_1) = 5x_1^2 - 8y_1$

$5(xx_1) - 4(y + 2) = 5 - 8 \cdot 2$

$5x - 4y + 3 = 0$

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