Question

Let $A B$ is the latus rectum of the hyperbola $\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1$ such that triangle $O A B$ is equilateral where ' $O$ ' is origin and under this condition eccentricity of the hyperbola is given as $\frac{1+\sqrt{p}}{2 \sqrt{q}}$ (where $p, q$ are prime numbers) then $p-q$ is

Answer 1

Since $O A=A B$ (equilateral triangle)
$\therefore a^{2} e^{2}+\frac{b^{4}}{a^{2}}=4 \cdot \frac{b^{4}}{a^{2}}$
$ \Rightarrow a^{2} e^{2}=3 \cdot \frac{b^{4}}{a^{2}}$
$\Rightarrow e^{2}=3 \cdot \frac{b^{4}}{a^{4}}$ ....(1)

Also $e^{2}=1+\frac{b^{2}}{a^{2}}$
$\Rightarrow e^{2}=1+\frac{e}{\sqrt{3}} $ (using (1))
$\Rightarrow \sqrt{3} e^{2}-e-\sqrt{3}=0 $
$\Rightarrow e=\frac{1+\sqrt{13}}{2 \sqrt{3}} $
$\therefore p-q=13-3=10$