Coordinate Geometry – Parabola – JEE Main 21 Jan 2026 Shift 1

Solution:

Let the vertex be $O(0, 0)$.
Any point $Q$ on the parabola $x^2 = 4y$ can be taken as $(2t, t^2)$.

Let $P(h, k)$ be the point dividing $OQ$ in the ratio $2:3$.
Using the section formula:
$$h = \frac{2(2t) + 3(0)}{2+3} = \frac{4t}{5} \Rightarrow t = \frac{5h}{4}$$
$$k = \frac{2(t^2) + 3(0)}{2+3} = \frac{2t^2}{5}$$

Eliminating $t$ to find the locus:
Substitute $t = \frac{5h}{4}$ into the expression for $k$:
$$k = \frac{2}{5} \left( \frac{5h}{4} \right)^2$$
$$k = \frac{2}{5} \left( \frac{25h^2}{16} \right)$$
$$k = \frac{5h^2}{8}$$
$$8k = 5h^2$$
Replacing $(h, k)$ with $(x, y)$, the equation of the conic $C$ is $5x^2 – 8y = 0$.

We need the equation of the chord of $C$ bisected at the point $(1, 2)$.
Using the formula $T = S_1$:
Here, $S \equiv 5x^2 – 8y$.
Point $(x_1, y_1) = (1, 2)$.

Calculate $T$:
$$T \equiv 5xx_1 – 4(y + y_1)$$
$$T \equiv 5x(1) – 4(y + 2) = 5x – 4y – 8$$

Calculate $S_1$:
$$S_1 \equiv 5(1)^2 – 8(2) = 5 – 16 = -11$$

Equating $T = S_1$:
$$5x – 4y – 8 = -11$$
$$5x – 4y + 3 = 0$$

Ans. (4)