Question

If $P Q$ is a double ordinate of 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 eccentricity $e$ of the hyperbola satisfies

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

Answer 1

Answer: (D)

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,
$P Q=2 b \tan \theta $
$C Q=C P=\sqrt{a^{2} \sec ^{2} \theta+b^{2} \tan ^{2} \theta} $
since $ C Q=C P=P Q$
$\therefore 4 b^{2} \tan ^{2} \theta=a^{2} \sec ^{2} \theta+b^{2} \tan ^{2} \theta$
$\Rightarrow 3 b^{2} \tan ^{2} \theta=a^{2} \sec ^{2} \theta$

$\Rightarrow 3 b^{2} \sin ^{2} \theta=a^{2}$
$\Rightarrow 3 a^{2}\left(e^{2}-1\right) \sin ^{2} \theta=a^{2}$
$\Rightarrow 3\left(e^{2}-1\right) \sin ^{2} \theta=1$
$\Rightarrow \frac{1}{3\left(e^{2}-1\right)}=\sin ^{2} \theta < 1$
$(\because \,\sin^2 \theta < 1 )$
$\Rightarrow \frac{1}{e^{2}-1} < 3 \Rightarrow e^{2}-1>\frac{1}{3}$