Q28 / 75 · JEE Main 2026 (21 January Shift 1) · Mathematics
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 :
- Correct answerA$5 x-4 y+3=0$
- B$5 x-y-3=0$
- C$4 x-5 y+6=0$
- D$x-2 y+3=0$
Solution
Parabola $x^2 = 4y$, vertex O = (0, 0), Q = $(2t, t^2)$ on parabola.
P divides OQ in ratio 2:3: $P = \left(\frac{4t}{5}, \frac{2t^2}{5}\right)$.
Eliminating $t$: $x^2 = \frac{8y}{5}$ (conic C with $a = \frac{2}{5}$).
For chord with midpoint $(h, k)$: $hx - 2ay = h^2 - 2ak$.
At $(1, 2)$: $x - \frac{4y}{5} = 1 - \frac{8}{5} = -\frac{3}{5}$.
$5x - 4y + 3 = 0$.
P divides OQ in ratio 2:3: $P = \left(\frac{4t}{5}, \frac{2t^2}{5}\right)$.
Eliminating $t$: $x^2 = \frac{8y}{5}$ (conic C with $a = \frac{2}{5}$).
For chord with midpoint $(h, k)$: $hx - 2ay = h^2 - 2ak$.
At $(1, 2)$: $x - \frac{4y}{5} = 1 - \frac{8}{5} = -\frac{3}{5}$.
$5x - 4y + 3 = 0$.
Hence, the correct answer is option: 'A'.