JEE Main · 2024 · 30 Jan 2024 · Shift 2 · Hyperbola

For $\frac{x^2}{9}-\frac{y^2}{4}=1$, $a^2=9$, $b^2=4$, so $e^2=1+\frac{4}{9}=\frac{13}{9}$.

The foci are at $(\pm ae,0)=(\pm \sqrt{13},0)$, a distance $2\sqrt{13}$ apart.

For $P=(x_0,y_0)$ on the hyperbola, the triangle formed with the two foci has area $\frac{1}{2}\cdot 2\sqrt{13}\cdot |y_0|=\sqrt{13}|y_0|$. Setting this equal to $2\sqrt{13}$ gives $|y_0|=2$.

Substituting $y_0=2$ into the hyperbola equation: $\frac{x_0^2}{9}-1=1\Rightarrow x_0^2=18$.

The square of the distance of $P$ from the origin is $x_0^2+y_0^2=18+4=22$.