Question

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

• 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

Answer: (D)

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

Let the hyperbola be $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$ and any double ordinate $PQ$ be $(a\sec \theta ,b\tan \theta ),(a\sec \theta ,-b\tan \theta )$ and $O$ is centre $(0,0)$.
$\Delta OPQ$ being equilateral.
Thus $\tan {30}^o=\frac{b\tan \theta }{a\sec \theta }$
$\Rightarrow 3\cdot \frac{b^2}{a^2}={\csc }^2\theta$
$\Rightarrow 3(e^2-1)={\csc }^2\theta$
Now, ${\csc }^2\theta \ge 1$
Therefore, $3(e^2-1)\ge 1\Rightarrow e^2\ge \frac{4}{3}$
$\Rightarrow e>2/\sqrt{3}$