JEE Main 2025 — Hyperbola Question with Solution

To find $p^2+q^2$ for the hyperbola $\mathrm{H}:\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$, given that the product of the focal distances from a point $\mathbf{P}(4,2\sqrt{3})$ to the foci is 32, we follow these steps:

Verify the Point on the Hyperbola:

The point $P(4,2\sqrt{3})$ satisfies the hyperbola equation:

$\frac{16}{a^2}-\frac{12}{b^2}=1\dots \text{(i)}$

Product of Focal Distances:

The distances from $P$ to the foci $S$ and $S^{\prime }$ are:

$SP=e\left(4-\frac{a}{e}\right),S^{\prime }P=e\left(4+\frac{a}{e}\right)$

Therefore, the product:

$SP\cdot S^{\prime }P=16e^2-a^2=32$

Relating $e$, $a$, and $b$:

Using the identity for eccentricity:

$e^2=1+\frac{b^2}{a^2}$

Substituting in the product equation gives:

$16\left(1+\frac{b^2}{a^2}\right)-a^2=32\dots \text{(ii)}$

Solving for $a^2$ and $b^2$:

From equations (i) and (ii), solve for $a^2$ and $b^2$:

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

Calculate $p^2+q^2$:

The conjugate axis length is $p=2b$ and the latus rectum is $q=\frac{2b^2}{a}$. Thus:

$p^2+q^2=(2b{)}^2+{\left(\frac{2b^2}{a}\right)}^2$

Substituting $b^2=12$ and $a^2=8$ gives:

$p^2+q^2=4b^2+\frac{4b^4}{a^2}=48+\frac{4\times 144}{8}$

$p^2+q^2=48\cdot \left(1+\frac{12}{8}\right)=120$

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