Question

A double ordinate $PQ$ of the hyperbola $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$ is such that $\Delta OPQ$ is equilateral, $O$ being the centre of the hyperbola. Then the eccentricity $e$ satisfies the relation

• A. $1<e<\frac{2}{\sqrt{3}}$
• B. $e=\frac{2}{\sqrt{3}}$
• C. $e=\frac{\sqrt{3}}{2}$
• D. $e>\frac{2}{\sqrt{3}}$

Answer 1

The correct option is D $e>\frac{2}{\sqrt{3}}$

Let any double ordinate $P,Q$ be $(a\sec \theta ,b\tan \theta )$ and $(a\sec \theta ,-b\tan \theta )$.

In $\Delta OPR$,
$\tan {30}^{\circ }=\frac{b\tan \theta }{a\sec \theta }$
$\Rightarrow \frac{1}{\sqrt{3}}=\frac{b}{a}\sin \theta$
$\Rightarrow \frac{b}{a}=\frac{1}{\sqrt{3}\sin \theta }$

$e^2=1+\frac{b^2}{a^2}$
$=1+\frac{1}{3{\sin }^2\theta }$
$\Rightarrow e^2>1+\frac{1}{3}(\because max[{\sin }^2\theta ]=1)$
$\Rightarrow e^2>\frac{4}{3}$
$\Rightarrow e>\frac{2}{\sqrt{3}}$