JEE Main 2024 — Hyperbola Question with Solution

Plotting the diagram of given data we get,

Given equation of hyperbola is $\frac{x^2}{9}-\frac{y^2}{4}=1$.

$\Rightarrow a^2=9, b^2=4$

We know that, $b^2=a^2\left(e^2-1\right)$

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

$\Rightarrow e^2=1+\frac{4}{9}$

$\Rightarrow e^2=\frac{13}{9}$

$\Rightarrow e=\frac{\sqrt{13}}{3}$

Now, $S_1{S}_2=2ae$

$\Rightarrow S_1{S}_2=2\times 3\times \sqrt{\frac{13}{3}}$

$\Rightarrow S_1{S}_2=2\sqrt{13}$

It is given that, area of $\Delta P{S}_1{S}_2$ is $2\sqrt{13}$.

$\Rightarrow \frac{1}{2}\times \beta \times S_1{S}_2=2\sqrt{13}$

$\Rightarrow \frac{1}{2}\times \beta \times 2\sqrt{13}=2\sqrt{13}$

$\Rightarrow \beta =2$

Since, $\left(\alpha ,\beta \right)$ lies on the hyperbola,

$\Rightarrow \frac{{\alpha }^2}{9}-\frac{{\beta }^2}{4}=1$

$\Rightarrow \frac{{\alpha }^2}{9}-1=1$

$\Rightarrow {\alpha }^2=18$

$\Rightarrow \alpha =3\sqrt{2}$

Distance of $P$ from origin is given by,

$OP=\sqrt{{\alpha }^2+{\beta }^2}$

$\Rightarrow OP=\sqrt{18+4}$

$\Rightarrow OP=\sqrt{22}$

$\Rightarrow O{P}^2=22$