Question

Let $AB$ is the latus rectum of the hyperbola $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$ such that triangle $OAB$ 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

Answer: 10

Since $OA=AB$ (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))
$\Rightarrow \sqrt{3}{e}^2-e-\sqrt{3}=0$
$\Rightarrow e=\frac{1+\sqrt{13}}{2\sqrt{3}}$
$\therefore p-q=13-3=10$