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

The Correct Option is C

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:

\[ x^2 + y^2 = 22. \]