Question

Let the latus rectum of the hyperbola \( \frac{x^2}{9} - \frac{y^2}{b^2} = 1 \) subtend an angle of \( \frac{\pi}{3} \) at the center of the hyperbola. If \( b^2 \) is equal to \( \frac{1}{m} \left( 1 + \sqrt{n} \right) \), where \( l \) and \( m \) are coprime numbers, then \( l^2 + m^2 + n^2 \) is equal to \(\_\_\_\_\_\_\_\_\_\).

Answer 1

Correct Answer: 182

Given the hyperbola \( \frac{x^2}{9} - \frac{y^2}{b^2} = 1 \) with latus rectum subtending \( 60^\circ \) at the center, we have: \[ \tan 30^\circ = \frac{b^2 / a}{ae} = \frac{b^2}{a^2 e} = \frac{1}{\sqrt{3}} \] 
This gives \( e = \frac{\sqrt{5}}{3} \).  Using \( e^2 = 1 + \frac{b^2}{a^2} \):
\[ b^2 = 3b^4 + 27 \Rightarrow b^4 - 3b^2 - 27 = 0 \] 
Solving, we get \( b^2 = \frac{1}{3}(1 + \sqrt{13}) \) with \( l = 2 \), \( m = 3 \), and \( n = 13 \).
Thus, \[ l^2 + m^2 + n^2 = 4 + 9 + 169 = 182 \]