Let the product of the focal distances of the point $\mathrm{P}(4,2 \sqrt{3})$ on the hyperbola $\mathrm{H}: \frac{\mathrm{x}^2}{\mathrm{a}^2}-\frac{\mathrm{y}^2}{\mathrm{~b}^2}=1$ be 32 .
Let the length of the conjugate axis of $H$ be $p$ and the length of its latus rectum be q . Then $\mathrm{p}^2+\mathrm{q}^2$ is equal to _______

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Aicharya · Accepted Answer

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

Let the product of the focal distances of the point $\mathrm{P}(4,2 \sqrt{3})$ on the hyperbola $\mathrm{H}:…

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