JEE Mathematics 2026 Question with Solution

Given: The parabola is $y^2 = 12x$ and the chord joining $P_1(x_1, y_1)$ and $P_2(x_2, y_2)$ subtends a right angle at the vertex.

Find: $x_1x_2 - y_1y_2$.

The given parabola is

$$y^2 = 12x$$

which is of the standard form $y^2 = 4ax$. Hence,

$$4a = 12 \Rightarrow a = 3$$

The vertex of the parabola is at the origin $O(0,0)$.

Step 1: Parametric coordinates of points on the parabola.

The parametric form of a point on the parabola $y^2 = 4ax$ is

$$(at^2,\, 2at)$$

Therefore, the coordinates of points $P_1$ and $P_2$ are

$$P_1(3t_1^2,\, 6t_1), \quad P_2(3t_2^2,\, 6t_2)$$

Step 2: Condition for right angle at the vertex.

Since the chord $P_1P_2$ subtends a right angle at the vertex $O$, we have

$$\overrightarrow{OP_1} \cdot \overrightarrow{OP_2} = 0$$

Thus,

$$(3t_1^2)(3t_2^2) + (6t_1)(6t_2) = 0$$ $$9t_1^2t_2^2 + 36t_1t_2 = 0$$

Dividing by $9$,

$$t_1^2t_2^2 + 4t_1t_2 = 0$$ $$t_1t_2(t_1t_2 + 4) = 0$$

Since the points are distinct,

$$t_1t_2 = -4$$

Step 3: Compute $x_1x_2 - y_1y_2$.

Using the parametric coordinates,

$$x_1x_2 = (3t_1^2)(3t_2^2) = 9t_1^2t_2^2$$ $$y_1y_2 = (6t_1)(6t_2) = 36t_1t_2$$

So,

$$x_1x_2 - y_1y_2 = 9t_1^2t_2^2 - 36t_1t_2$$

Substituting $t_1t_2 = -4$,

$$x_1x_2 - y_1y_2 = 9(16) - 36(-4)$$ $$= 144 + 144 = 288$$

Therefore, the correct option is B.

Given: $y^2 = 12x = 4ax$, so $a = 3$.

Find: $x_1x_2 - y_1y_2$.

For a parabola $y^2 = 4ax$, if the chord joining parameter points $t_1$ and $t_2$ subtends a right angle at the vertex, then

$$t_1t_2 = -4$$

Now,

$$x_1 = at_1^2, \quad x_2 = at_2^2, \quad y_1 = 2at_1, \quad y_2 = 2at_2$$

Hence,

$$x_1x_2 - y_1y_2 = a^2 t_1^2 t_2^2 - 4a^2 t_1t_2$$

With $a = 3$,

$$x_1x_2 - y_1y_2 = 9(t_1t_2)^2 - 36(t_1t_2)$$

Substituting $t_1t_2 = -4$,

$$x_1x_2 - y_1y_2 = 9(16) + 144 = 288$$

Therefore, the correct option is B.