JEE Main 2025 — Hyperbola Question with Solution

$\frac{x^2}{a^2}-\frac{y^2}{b^2}=1....(1)$
$\begin{array}{rl}& \mathrm{P}(4,2\sqrt{3})\\ & {\mathrm{PS}}_1\cdot {\mathrm{PS}}_2=32\\ & \left|{\mathrm{PS}}_1-{\mathrm{PS}}_2\right|=2\mathrm{a}\\ & \mathrm{P}(4,2\sqrt{3})\text{ lies on }\mathrm{H}\\ & \therefore \frac{16}{{\mathrm{a}}^2}-\frac{12}{{\mathrm{b}}^2}=1\end{array}$
$16b^2-12a^2=a^2{b}^2....(2)$
$\begin{aligned}
& \left|\mathrm{PS}_1-\mathrm{PS}_2\right|^2=4 \mathrm{a}^2 \\ & \mathrm{PS}_1^2+\mathrm{PS}_2{ }^2-2 \mathrm{PS}_1 \cdot \mathrm{PS}_2=4 \mathrm{a}^2 \\ & (\mathrm{ae}-4)^2+12+(\mathrm{ae}+4)^2+12-64=4 \mathrm{a}^2 \\ & 2 \mathrm{a}^2 \mathrm{e}^2-8=4 \mathrm{a}^2 \\ & \mathrm{a}^2+\mathrm{b}^2-4=2 \mathrm{a}^2 \\ & \mathrm{~b}^2-\mathrm{a}^2=4
\end{aligned}$<br/>$\begin{aligned}
& (2) \&(3) \Rightarrow 16\left(a^2+4\right)-12 a^2=a^2\left(a^2+4\right) \\ & \Rightarrow 16 a^2+64-12 a^2=a^4+4 a^2 \\ & \Rightarrow a^4=64 \\ & \Rightarrow a^2=8
\end{aligned}$<br/>$\begin{aligned} & \therefore \mathrm{b}^2=12 \\ & \mathrm{p}^2+\mathrm{q}^2=4 \mathrm{~b}^2+\frac{4 \mathrm{~b}^4}{\mathrm{a}^2} \\ & =120\end{aligned}$