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