JEE Mathematics 2026 Question with Solution

Given: The hyperbola is

$$\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$$

and the point $P(10, 2\sqrt{15})$ lies on it. The length of the latus rectum is $8$.

Find: The square of the area of $\triangle PSS'$, where $S$ and $S'$ are the foci.

For the hyperbola, the latus rectum length is

$$\frac{2b^2}{a} = 8$$

So,

$$b^2 = 4a$$

Using the point $P(10, 2\sqrt{15})$ in the hyperbola equation,

$$\frac{100}{a^2} - \frac{60}{b^2} = 1$$

Substituting $b^2 = 4a$,

$$\frac{100}{a^2} - \frac{60}{4a} = 1$$ $$\frac{100}{a^2} - \frac{15}{a} = 1$$

Multiplying by $a^2$,

$$100 - 15a = a^2$$ $$a^2 + 15a - 100 = 0$$

Hence,

$$a = 5, \qquad b^2 = 20$$

Now for the hyperbola,

$$c^2 = a^2 + b^2 = 25 + 20 = 45$$

Therefore,

$$c = 3\sqrt{5}$$

So the distance between the foci is

$$SS' = 2c = 6\sqrt{5}$$

The base $SS'$ lies on the $x$-axis and the height of point $P$ from the $x$-axis is $2\sqrt{15}$. Therefore,

$$\text{Area} = \frac{1}{2} \times 6\sqrt{5} \times 2\sqrt{15}$$ $$\text{Area} = 6\sqrt{75} = 30\sqrt{3}$$

Hence,

$$(\text{Area})^2 = (30\sqrt{3})^2 = 2700$$

Therefore, the correct option is D.

Given: $P(10, 2\sqrt{15})$ lies on $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$ and the latus rectum length is $8$.

Find: $\left(\text{Area of } \triangle PSS'\right)^2$.

First use the latus rectum formula:

$$\frac{2b^2}{a} = 8$$ $$b^2 = 4a$$

Now substitute the coordinates of $P$:

$$\frac{10^2}{a^2} - \frac{(2\sqrt{15})^2}{b^2} = 1$$ $$\frac{100}{a^2} - \frac{60}{b^2} = 1$$

Using $b^2 = 4a$,

$$\frac{100}{a^2} - \frac{60}{4a} = 1$$ $$\frac{100}{a^2} - \frac{15}{a} = 1$$

Multiply throughout by $a^2$:

$$100 - 15a = a^2$$ $$a^2 + 15a - 100 = 0$$

Factoring gives the valid positive value

$$a = 5$$

Then,

$$b^2 = 4a = 20$$

For the foci of the hyperbola,

$$c^2 = a^2 + b^2$$ $$c^2 = 25 + 20 = 45$$ $$c = 3\sqrt{5}$$

Hence the distance between the foci is

$$2c = 6\sqrt{5}$$

Now take $SS'$ as the base of the triangle. Since the foci lie on the transverse axis, $SS'$ is on the $x$-axis. The ordinate of $P$ is $2\sqrt{15}$, so the perpendicular height is $2\sqrt{15}$.

Thus,

$$\text{Area of } \triangle PSS' = \frac{1}{2} \times 6\sqrt{5} \times 2\sqrt{15}$$ $$= 6\sqrt{75}$$ $$= 30\sqrt{3}$$

Now square the area:

$$(30\sqrt{3})^2 = 2700$$

Therefore, the square of the area is $2700$, so the correct option is D.