Question

$PQ$ is a double ordinate of the hyperbola $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$ such that $OPQ$ is equilateral triangle, $O$ being the centre of the hyperbola, then find range of the eccentricity $e$ of the hyperbola.

Answer 1

$\angle POR=30$

$\tan {30}^0=PR/OR$

$\frac{1}{\sqrt{3}}=\frac{asec\theta }{b\tan \theta }$

$\frac{1}{\sqrt{3}}=\frac{a\cos \theta }{b\sin \theta \cos \theta }\Rightarrow \frac{1}{\sqrt{3}}=\frac{a}{bsin\theta }$

$\frac{b}{a\sqrt{3}}=\frac{1}{sin\theta }$ $b^2=a^2(e^2-1)$

$cosec\theta =\frac{b}{a}\sqrt{3}$ $b^2/a^2=e^2-1$......(1)

$cose{c}^2\theta =3b^2/a^2$

$cose{c}^2\theta =3(e^2-1)$

Now,

$cose{c}^2\theta \ge 1$

$3(e^2-1)\ge 1$

$e^2-1\ge \frac{1}{3}$

$e^2\ge 1+\frac{1}{3}\Rightarrow e^2\ge \frac{4}{3}$

$\Rightarrow e\ge \frac{2}{\sqrt{3}}\Rightarrow e>\frac{2}{\sqrt{3}}$