Question

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

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

Let $P(a\sec \theta ,b\tan \theta );Q(a\sec \theta ,-b\tan \theta )$ be end points of double ordinate and $C(0,0)$ is the centre of the hyperbola.

Now, $PQ=2b\tan \theta$

$CQ=CP=\sqrt{a^2{\sec }^2\theta +b^2{\tan }^2\theta }$

since $CQ=CP=PQ$

$\therefore 4b^2{\tan }^2\theta =a^2{\sec }^2\theta +b^2{\tan }^2\theta \Rightarrow 3b^2{\tan }^2\theta =a^2{\sec }^2\theta$

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

$3(e^2-1){\sin }^2\theta =1\Rightarrow \frac{1}{3(e^2-1)}={\sin }^2\theta <1$ $(\because {\sin }^2\theta <1)$

$\Rightarrow \frac{1}{e^2-1}<3\Rightarrow e^2-1>\frac{1}{3}$