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 centre of the hyperbola, then the eccentricity ${}^{\prime }{e}^{\prime }$ 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 $P(a\sec \theta ,b\tan \theta ),Q(a\sec \theta ,-b\tan \theta )$ be end points of double ordinates and $(0,0)$ is the centre of the hyperbola.
So, $PQ=2b\tan \theta$
$OQ=OP=\sqrt{a^2{\sec }^2\theta +b^2{\tan }^2\theta }$
Since, $OQ=OP=PQ$
$\therefore 4b^2{\tan }^2\theta =a^2{\sec }^2\theta +b^2{\tan }^2\theta$
$\Rightarrow 3b^2{\tan }^2\theta =a^2{\sec }^2\theta$
$\Rightarrow 3b^2{\sin }^2\theta =a^2$
$\Rightarrow 3a^2(e^2-1){\sin }^2\theta =a^2$
$\Rightarrow 3(e^2-1){\sin }^2\theta =1$
$\Rightarrow \frac{1}{3(e^2-1)}={\sin }^2\theta <1.(\because {\sin }^2\theta <1)$
$\Rightarrow \frac{1}{e^2-1}<3\Rightarrow e^2-1>\frac{1}{3}\Rightarrow e^2>\frac{4}{3}$
$\therefore e>\frac{2}{\sqrt{3}}$