Coordinate Geometry – Parabola – JEE Main 22 Jan 2026 Shift 1

Solution:

For the parabola $y^2 = 12x$, we have $4a = 12 \Rightarrow a = 3$.
Let the points on the parabola be in parametric form $P(at^2, 2at)$.
$$ P_1(x_1, y_1) = (3t_1^2, 6t_1) $$
$$ P_2(x_2, y_2) = (3t_2^2, 6t_2) $$

Since the chord $P_1P_2$ subtends a right angle at the vertex $(0,0)$, the product of the slopes of $OP_1$ and $OP_2$ is $-1$.
$$ m_{OP_1} \cdot m_{OP_2} = -1 $$
$$ \left(\frac{6t_1}{3t_1^2}\right) \cdot \left(\frac{6t_2}{3t_2^2}\right) = -1 $$
$$ \left(\frac{2}{t_1}\right) \cdot \left(\frac{2}{t_2}\right) = -1 $$
$$ \frac{4}{t_1t_2} = -1 \Rightarrow t_1t_2 = -4 $$

We need to find the value of $x_1x_2 – y_1y_2$.
$$ x_1x_2 = (3t_1^2)(3t_2^2) = 9(t_1t_2)^2 $$
$$ y_1y_2 = (6t_1)(6t_2) = 36(t_1t_2) $$

Substitute $t_1t_2 = -4$:
$$ x_1x_2 = 9(-4)^2 = 9(16) = 144 $$
$$ y_1y_2 = 36(-4) = -144 $$

Now calculate the expression:
$$ x_1x_2 – y_1y_2 = 144 – (-144) $$
$$ = 144 + 144 = 288 $$

Ans. (1)