Question

If the latus rectum of a hyperbola forms an equilateral triangle with the vertex at the center of the hyperbola, then its eccentricity $e=$?

• A. $\frac{\sqrt{5}+1}{2}$
• B. $\frac{\sqrt{11}+1}{2}$
• C. $\frac{\sqrt{13}+1}{2\sqrt{3}}$
• D. $\frac{\sqrt{13}-1}{2\sqrt{3}}$

Answer 1

Answer: (C)

$\frac{\sqrt{13}+1}{2\sqrt{3}}$

Ends of latus rectum

$\begin{array}{c}\tan {30}^{\circ }=\frac{\frac{b^2}{a}}{ae}\\ \frac{1}{\sqrt{3}}=\frac{b^2}{a^{2}e}=\frac{e^2-1}{e}\\ \sqrt{3}{e}^2-e-\sqrt{3}=0\\ e=\frac{1\pm \sqrt{1+12}}{2\sqrt{3}}=\frac{\sqrt{13}+1}{2\sqrt{3}}\end{array}$