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

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, $l^2 + m^2 + n^2 = 4 + 9 + 169 = 182$