JEE Mathematics 2025 Question with Solution

Given: Hyperbola $H: \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$, point $P(4, 2\sqrt{3})$ lies on it, and $PS_1 \cdot PS_2 = 32$.

Find: $p^2 + q^2$, where $p$ is the length of the conjugate axis and $q$ is the length of the latus rectum.

Since $P(4, 2\sqrt{3})$ lies on the hyperbola,

So,

Also, for a point on the hyperbola, the absolute difference of focal distances is $2a$:

Hence,

Using

and the coordinates of $P$,

Since

we get

Therefore,

Now use equations $(2)$ and $(3)$. From $b^2 = a^2 + 4$,

Thus,

The length of the conjugate axis is

and the length of the latus rectum is

So,

Substituting $a^2 = 8$ and $b^2 = 12$,

Therefore, $p^2 + q^2 = 120$.

Given: $P(4, 2\sqrt{3})$ lies on $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$ and the product of its focal distances is $32$.

Find: $p^2 + q^2$.

Let the foci be $S_1(-c,0)$ and $S_2(c,0)$, where

The focal distances are

Their product is given as $32$, so

Squaring both sides,

Put $u = c^2$. Then

So the positive value is

Hence,

Since $P(4, 2\sqrt{3})$ lies on the hyperbola,

Using this with $a^2 + b^2 = 20$ gives

Now,

Therefore,

Therefore, the required value is $120$.