The tangent at any point P on the ellipse $\frac{{\mathrm{x}}^2}{{\mathrm{a}}^2}+\frac{{\mathrm{y}}^2}{{\mathrm{b}}^2}=1$ makes an angle $\mathrm{\alpha }=\frac{5\mathrm{\pi }}{6}$ with the positive x – axis and the angle subtended by the foci at the point P is $\mathrm{\beta }=\frac{\mathrm{\pi }}{3}$. If eccentricity of the ellipse is $e$, then $\frac{3}{{\mathrm{e}}^2}$ is equal to …...

Let angle S₁ PT be $\mathrm{\gamma }$
By the reflection property of ellipse $\mathrm{\gamma }=\frac{\mathrm{\pi }}{2}-\frac{\mathrm{\beta }}{2}$
Equation of tangent at P is $\frac{\mathrm{x}}{\mathrm{a}}\cos \mathrm{\theta }+\frac{\mathrm{y}}{\mathrm{b}}\sin \mathrm{\theta }=1$
Therefore slope $\tan \mathrm{\alpha }=\frac{-\mathrm{b}}{\mathrm{a}}\cot \mathrm{\theta }\dots .\left(1\right)$
Apply sine rule in triangle S₁ PT
$\frac{{\mathrm{PS}}_1}{\sin (\mathrm{\pi }-\mathrm{\alpha })}=\frac{{\mathrm{TS}}_1}{\sin \mathrm{\gamma }}=>\frac{\mathrm{a}-\mathrm{eacos}\mathrm{\varphi }}{\sin \mathrm{\alpha }}=\frac{{\cos }^{\mathrm{a}}\mathrm{ae}}{\sin \mathrm{\gamma }}$
Simplifying it we get 
$$\begin{gathered} \text{cos⁡ϕ=}\frac{\sin \mathrm{\alpha }}{\sin \mathrm{\gamma }}\dots \dots ..\left(2\right) \\ \mathrm{Also} \mathrm{l} -{\mathrm{e}}^2=\frac{{\mathrm{b}}^2}{{\mathrm{a}}^2}\dots \dots \left(3\right) \end{gathered}$$
Using (1), (2) and (3) eliminating $\mathrm{\varphi }$ We get, $\mathrm{e}=\frac{\cos \mathrm{\gamma }}{\cos \mathrm{x}}=-\frac{\sin \mathrm{\beta }/2}{\cos \mathrm{\alpha }}$

$e=-\frac{\sin 30°}{\cos 150°}=\frac{1}{\sqrt{3}}$

$\Rightarrow \frac{3}{{\mathrm{e}}^2}=9$