Question

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

• A.
  
• B.
  
• C.
  
• D.

Answer 1

The correct option is D

It is given that $\Delta A^{{}^{\prime }}L{L}^{{}^{\prime }}$is an equilateral triangle. Therefore, $\angle L{A}^{{}^{\prime }}S={30}^{\circ }$
In $\Delta A^{{}^{\prime }}$ LS,we have tan $\angle L{A}^{{}^{\prime }}S=\frac{LS}{A^{{}^{\prime }}S}$
$\Rightarrow$ $tan{30}^{\circ }=\frac{b^2/a}{a+ae}$
$\Rightarrow$ $\frac{1}{\sqrt{3}}=\frac{b^2}{a^2(1+e)}$
$\Rightarrow$ $\frac{1}{\sqrt{3}}=\frac{e^2-1}{1+e}$
$\Rightarrow$ $\frac{1}{\sqrt{3}}=e-1$
$\Rightarrow$ $e=1+\frac{1}{\sqrt{3}}=\frac{\sqrt{3}+1}{\sqrt{3}}$