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

from triangle $\mathrm{Cs}=\mathrm{ae}, \mathrm{SL}=\frac{{\mathrm{b}}^2}{\mathrm{a}}$

from $∆\mathrm{le} \mathrm{CSL},$

$$\begin{gathered} \tan {30}^0=\frac{{\mathrm{b}}^2}{{\displaystyle \frac{\mathrm{a}}{\mathrm{ae}}}} \\ \Rightarrow \frac{1}{\sqrt{3}}=\frac{{\mathrm{b}}^2}{{\mathrm{a}}^2\mathrm{e}}=\frac{\cancel{{\mathrm{a}}^2} \left({\mathrm{e}}^2-1\right)}{\cancel{{\mathrm{a}}^2}\mathrm{e}} \\ \Rightarrow \sqrt{3}\left({\mathrm{e}}^2-1\right)=\mathrm{e} \\ \Rightarrow \sqrt{3}{\mathrm{e}}^2-\mathrm{e}-\sqrt{3}=0 \\ \Rightarrow \mathrm{e}=\frac{1\pm \sqrt{1-4\left(\sqrt{3}\right)(-\sqrt{3})}}{2\left(\sqrt{3}\right)} \\ =\frac{1\pm \sqrt{13}}{2\sqrt{3}} \\ \Rightarrow \mathrm{e}=\frac{\sqrt{13}+1}{2\sqrt{3}} (\therefore \mathrm{e} > 0) \end{gathered}$$