Q. A double ordinate $PQ$ of the hyperbola $\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1$ is such that $\triangle O P Q$ is equilateral, $O$ being the centre of the hyperbola. Then the eccentricity e satisfies the relation
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Solution:
Let, $P =( a \sec \theta, b \tan \theta) \angle POM =\tan 30^{\circ}=\frac{1}{\sqrt{3}}=\frac{b}{a} \sin \theta$
$\Rightarrow \sin \theta \frac{a}{b \sqrt{3}}$
$\Rightarrow \frac{b^{2}}{a^{2}}>\frac{1}{3}$
$\Rightarrow 1+\frac{b^{2}}{a^{2}}>\frac{4}{3}$
$\Rightarrow e>\frac{2}{\sqrt{3}}$
