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

Let the equation of the hyperbola be 

$\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$ 

$F_1(-ae,0),F_2(ae,0)$ be the foci, $e$ being the eccentricity.

Let $L{F}_2{L}^{\mathrm{\prime }},$ the latus rectum subtend an angle of 60° at $F_1,$ 

then slope of $L{F}_1$ is $\tan {30}^0.$

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

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