If the extremities of a focal chord are $\frac{5\mathrm{\pi }}{12}$ and $\frac{-\mathrm{\pi }}{12}$ then $\mathrm{e}=$

$$\begin{gathered} \mathrm{If} \mathrm{\alpha } \mathrm{and} \mathrm{\beta } \mathrm{are} \mathrm{extremities} \mathrm{of} \mathrm{a} \mathrm{focal} \mathrm{chord} \mathrm{of} \mathrm{ellipse} \mathrm{then} \\ \mathrm{e}=\frac{\cos \left({\displaystyle \frac{\mathrm{\alpha }-\mathrm{\beta }}{2}}\right)}{\cos \left(\frac{\mathrm{\alpha }+\mathrm{\beta }}{2}\right)} \\ where \mathrm{\alpha }=\frac{5\mathrm{\pi }}{12} , \mathrm{\beta }=\frac{\pi }{12} \\ \mathrm{e}=\frac{\cos \left({\displaystyle \raisebox{1ex}{$\mathrm{\pi }$}\!\left/ \!\raisebox{-1ex}{$4$}\right.}\right)}{\cos \left(\mathrm{\pi }/6\right)}=\frac{{\displaystyle \frac{1}{\sqrt{2}}}}{{\displaystyle \frac{\sqrt{3}}{2}}}=\sqrt{\frac{2}{3}} \end{gathered}$$