JEE Main 2025 Apr 3 Shift 1, Mathematics Q22: Introduction to Hyperbola

Given,

$\frac{x^2}{a^2}-\frac{y^2}{b^2}=1,P(4,2\sqrt{3})$ lies on the hyperbola.

$\therefore \frac{16}{a^2}-\frac{12}{b^2}=1$

$\Rightarrow 16b^2-12a^2=a^2{b}^2...(1)$

Also, $|P{S}_1-P{S}_2|=2a$ and $P{S}_1\cdot P{S}_2=32$.

Since the foci are $(\pm ae,0)$,

$P{S}_1^2=(ae-4{)}^2+12,P{S}_2^2=(ae+4{)}^2+12$

$\Rightarrow (P{S}_1-P{S}_2{)}^2=4a^2$

$\Rightarrow (ae-4{)}^2+12+(ae+4{)}^2+12-64=4a^2$

$\Rightarrow e^2=4$

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

$\Rightarrow b^2=3a^2$

Substituting in (1),

$48a^2-12a^2=3a^4\Rightarrow a^2=8,b^2=12$

$p=2b,q=\frac{2b^2}{a}$

$\therefore p^2+q^2=4b^2+\frac{4b^4}{a^2}=48+72=\boxed{120}$