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A double ordinate $\mathrm{PQ}$ of the hyperbola $\frac{\mathrm{x}^{2}}{\mathrm{a}^{2}}-\frac{\mathrm{y}^{2}}{\mathrm{~b}^{2}}=1$ is such that $\Delta \mathrm{OPQ}$ is equilateral, $\mathrm{O}$ being the centre of the hyperbola. Then the eccentricity e satisfies the relation
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Verified Answer
The correct answer is:
$\mathrm{e}>\frac{2}{\sqrt{3}}$
Hint:
$P \equiv(a \sec \theta, b \tan \theta) \angle \mathrm{POM}=\tan 30^{\circ}=\frac{1}{\sqrt{3}}=\frac{\mathrm{b}}{\mathrm{a}} \sin \theta$
$\Rightarrow \sin \theta=\frac{a}{b \sqrt{3}}$
$\Rightarrow \frac{b^{2}}{a^{2}}>\frac{1}{3} \quad \Rightarrow 1+\frac{b^{2}}{a^{2}}>\frac{4}{3} \quad \Rightarrow e>\frac{2}{\sqrt{3}}$
$P \equiv(a \sec \theta, b \tan \theta) \angle \mathrm{POM}=\tan 30^{\circ}=\frac{1}{\sqrt{3}}=\frac{\mathrm{b}}{\mathrm{a}} \sin \theta$
$\Rightarrow \sin \theta=\frac{a}{b \sqrt{3}}$
$\Rightarrow \frac{b^{2}}{a^{2}}>\frac{1}{3} \quad \Rightarrow 1+\frac{b^{2}}{a^{2}}>\frac{4}{3} \quad \Rightarrow e>\frac{2}{\sqrt{3}}$
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