Question

If $P Q$ is a double ordinate of the hyperbola $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$ such that $O P Q$ is an equilateral triangle, $O$ being the centre of the hyperbola, then the eccentricitye 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)

Let the hyperbola $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$ and any double ordinate $P Q$ be ( $\left.a \sec \theta, b \tan \theta\right)$, ( $\sec \theta,-b$ $\tan \theta) \cdot O$ is the centre $(0,0)$.
$\triangle O P Q$ is equilateral.

So $\tan 30^{\circ}=\frac{ b \tan \theta}{ a \sec \theta} $
$\Rightarrow 3 \frac{ b ^2}{ a ^2}=\operatorname{cosec}^2 \theta$
$\Rightarrow 3\left( e ^2-1\right)=\operatorname{cosec}^2 \theta $
$\text { Now } \operatorname{cosec}^2 \theta \geq 1$
Now $\operatorname{cosec}^2 \theta \geq 1$
$\Rightarrow 3\left( e ^2-1\right) \geq 1 \Rightarrow e ^2 \geq \frac{4}{3} \Rightarrow e >\frac{2}{\sqrt{3}} $