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

To solve this problem, follow these steps:

1. Identify the characteristics of the given hyperbola \( H: \frac{x^2}{9} - \frac{y^2}{4} = 1 \). It is centered at the origin with semi-axes lengths \( a = 3 \) and \( b = 2 \).
2. Calculate the distance to the foci. The distance to each focus \( c \) is given by \( c = \sqrt{a^2 + b^2} \). Therefore, \( c = \sqrt{9 + 4} = \sqrt{13} \). Thus, the foci are located at \( (\pm \sqrt{13}, 0) \).
3. Consider the point \( P(x, y) \) on the hyperbola such that its coordinates satisfy the hyperbola equation: \(\frac{x^2}{9} - \frac{y^2}{4} = 1\).
4. The area of the triangle formed by point \( P(x, y) \) and the foci \((\sqrt{13}, 0)\) and \((- \sqrt{13}, 0)\) is \( 2\sqrt{13} \). Use the area formula for a triangle with vertices \((x_1, y_1)\), \((x_2, y_2)\), and \((x_3, y_3)\):

\(\text{Area} = \frac{1}{2} \left| x_1(y_2-y_3) + x_2(y_3-y_1) + x_3(y_1-y_2) \right|\)

1. Substitute the coordinates: \((x_1, y_1) = (x, y)\), \((x_2, y_2) = (\sqrt{13}, 0)\), and \((x_3, y_3) = (-\sqrt{13}, 0)\):

\(\text{Area} = \frac{1}{2} \left| x(0 - 0) + \sqrt{13}(0 - y) - \sqrt{13}(y - 0) \right| = \frac{1}{2} \left| -2\sqrt{13}y \right| = \sqrt{13}|y|\)

1. Set the area equal to \( 2\sqrt{13} \):

\(\sqrt{13}|y| = 2\sqrt{13} \Rightarrow |y| = 2 \Rightarrow y = 2\)

1. Substitute \( y = 2 \) back into the hyperbola equation to find \( x \):

\(\frac{x^2}{9} - \frac{2^2}{4} = 1 \Rightarrow \frac{x^2}{9} - 1 = 1 \Rightarrow \frac{x^2}{9} = 2 \Rightarrow x^2 = 18 \Rightarrow x = \sqrt{18}\)

1. The point \( P \) is \( (\sqrt{18}, 2) \) or \( (3\sqrt{2}, 2) \). The distance of \( P \) from the origin is computed as follows:

\(\text{Distance from the origin} = \sqrt{x^2 + y^2} = \sqrt{18 + 4} = \sqrt{22}\)

1. Thus, the square of the distance of \( P \) from the origin is \( 22 \).

Therefore, the correct answer is 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:

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