Let LL' be the latus rectum through the focus of the hyperbola $\frac{x^{2}}{a^{2}} - \frac{y^{2}}{b^{2}} = 1$ and, A' be the farther vertex. If $△ A^{'} {LL}^{'}$ is equilateral, then the eccentricity of the hyperbola is

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$(\sqrt{3} + 1) / \sqrt{2}$

$(\sqrt{3} + 1) / \sqrt{3}$

$\sqrt{3} + 1$

$\sqrt{3}$

answer is D.

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Detailed Solution

$\begin{aligned} \tan 30^{\circ} = \frac{b^{2} / a}{a + ae} \\ ⇒ \frac{1 + e}{\sqrt{3}} = e^{2} - 1 \\ ⇒ e - 1 = \frac{1}{\sqrt{3}} \\ ⇒ e = \frac{\sqrt{3} + 1}{\sqrt{3}} \end{aligned}$

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