If the latus rectum of a hyperbola subtend an angle of 60° at the other focus, then eccentricity of the hyperbola is

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If the latus rectum of a hyperbola subtend an angle of 60° at the other focus, then eccentricity of the hyperbola is

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$\frac{\sqrt{3} + 1}{2}$

$\sqrt{3}$

$2 \sqrt{3}$

answer is D.

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Detailed Solution

Let the equation of the hyperbola be

$\frac{x^{2}}{a^{2}} - \frac{y^{2}}{b^{2}} = 1$

$F_{1} (- a e, 0), F_{2} (a e, 0)$ be the foci, $e$ being the eccentricity.

Let $L F_{2} L^{'},$ the latus rectum subtend an angle of 60° at $F_{1},$

then slope of $L F_{1}$ is $\tan 30^{0} .$

So $\tan 30^{\circ} = \frac{b^{2} / a}{2 a e}$

$\begin{array}{l} \Rightarrow \frac{1}{\sqrt{3}} = \frac{a^{2} (e^{2} - 1)}{2 a^{2} e} = \frac{e^{2} - 1}{2 e} \\ \Rightarrow e^{2} - 1 = \frac{2}{\sqrt{3}} e \Rightarrow {(e - \frac{1}{\sqrt{3}})}^{2} = 1 + \frac{1}{3} \\ \Rightarrow e - \frac{1}{\sqrt{3}} = \frac{2}{\sqrt{3}} \Rightarrow e = \sqrt{3} \end{array}$