Write the hyperbola and its foci

Given $H: \frac{x^2}{9}-\frac{y^2}{4}=1$

This is of the form $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$ with $a^2=9,\quad b^2=4$

For such a hyperbola, $c^2=a^2+b^2=9+4=13 \implies c=\sqrt{13}$

Hence the two foci are $F_1=(\sqrt{13},0),\qquad F_2=(-\sqrt{13},0)$

Use the area condition

Let $P=(x,y)$ be a point on the hyperbola in the first quadrant, so $x>0,y>0$.

The triangle is formed by $P$ and the two foci. The base is the segment joining the foci: $F_1F_2 = 2\sqrt{13}$

Since the foci lie on the $x$-axis, the perpendicular distance of $P$ from the line $F_1F_2$ is simply $y$.

Therefore, area of triangle $PF_1F_2$ is $\frac{1}{2}\cdot 2\sqrt{13}\cdot y = \sqrt{13}\,y$

Given area is $2\sqrt{13}$, so $\sqrt{13}\,y = 2\sqrt{13} \implies y=2$

Use the fact that $P$ lies on the hyperbola

Substitute $y=2$ into $\frac{x^2}{9}-\frac{y^2}{4}=1$

$\frac{x^2}{9}-\frac{4}{4}=1$ $\frac{x^2}{9}-1=1$ $\frac{x^2}{9}=2$ $x^2=18$

Find the square of the distance from the origin

The square of the distance of $P$ from the origin is $OP^2=x^2+y^2=18+4=22$

Check options

$22$ corresponds to Option B.

Compare with stored answer

Stored correct answer is B, which matches our result.