Question

If the latus rectum of a hyperbola forms an equilateral triangle with the vertex at the centre of the hyperbola, then eccentricity of the hyperbola is

• A.
• B.
• C.
• D.

Answer 1

The correct option is C

Let LSL’ be a latus rectum and C be the centre of the hyperbola $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$
It is given that $\Delta$ CLL’ is an equilateral triangle. Therefore, \(\angle LCS=30^\circ\)
In $\Delta$ CSL, we have $tan{30}^{\circ }=\frac{SL}{CS}$

$\Rightarrow \frac{1}{\sqrt{3}}=\frac{\frac{b^2}{2}}{ae}\Rightarrow \frac{1}{\sqrt{3}}=\frac{b^2}{a^{2}e}\Rightarrow \frac{1}{\sqrt{3}}=\frac{e^2-1}{e}\Rightarrow \sqrt{3}{e}^2-e-\sqrt{3}=0\Rightarrow e=\frac{1+\sqrt{13}}{2\sqrt{3}}$