Question

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

• A. 18
• B. 26
• C. 22
• D. 20

Answer 1

Answer: (C)

22

Answer 2

For the hyperbola $\frac{x^2}{9} - \frac{y^2}{4} = 1$, we have $a = 3$, $b = 2$, and $c = \sqrt{13}$, so the foci are at $\left(\pm \sqrt{13}, 0\right)$.

Let $P = (x, y) = \left(3 \sec \theta, 2 \tan \theta\right)$.

Given the area of the triangle with vertices at $P$ and the foci is $2\sqrt{13}$, we find that $\tan \theta = 1$, so $\theta = \frac{\pi}{4}$.

Substitute $\theta = \frac{\pi}{4}$:

$x = 3\sqrt{2}, \quad y = 2.$

The square of the distance from $P$ to the origin is:

$x^2 + y^2 = 22.$