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Let $\mathrm{P}$ be a point on the hyperbola $\mathrm{H}: \frac{\mathrm{x}^2}{9}-\frac{\mathrm{y}^2}{4}=1$, in the first quadrant such that the area of triangle formed by $\mathrm{P}$ and the two foci of $\mathrm{H}$ is $2 \sqrt{13}$. Then, the square of the distance of $\mathrm{P}$ from the origin is
$18$
$26$
$22$
$20$
Solution

$\frac{\mathrm{x}^2}{9}-\frac{\mathrm{y}^2}{4}=1$
$\mathrm{a}^2=9, \mathrm{~b}^2=4$
$\mathrm{~b}^2=\mathrm{a}^2\left(\mathrm{e}^2-1\right) \Rightarrow \mathrm{e}^2=1+\frac{\mathrm{b}^2}{\mathrm{a}^2}$
$\mathrm{e}^2=1+\frac{4}{9}=\frac{13}{9} $
$\mathrm{e}=\frac{\sqrt{13}}{3} \Rightarrow \mathrm{s}_1 \mathrm{~s}_2=2 \mathrm{ae}=2 \times 3 \times \sqrt{\frac{13}{3}}=2 \sqrt{13}$
Area of $\Delta \mathrm{PS}_1 \mathrm{~S}_2=\frac{1}{2} \times \beta \times \mathrm{s}_1 \mathrm{~S}_2=2 \sqrt{13}$
$\Rightarrow \frac{1}{2} \times \beta \times(2 \sqrt{13})=2 \sqrt{13} \Rightarrow \beta=2 $
$\frac{\alpha^2}{9}-\frac{\beta^2}{4}=1 \Rightarrow \frac{\alpha^2}{9}-1=1 \Rightarrow \alpha^2=18 \Rightarrow \alpha=3 \sqrt{2}$
Distance of $\mathrm{P}$ from origin $=\sqrt{\alpha^2+\beta^2}$
$=\sqrt{18+4}=\sqrt{22}$