If $\sqrt{3}\mathrm{bx}+\mathrm{ay}=2\mathrm{ab}$ is a tangent to the ellipse $\frac{{\mathrm{x}}^2}{{\mathrm{a}}^2}+\frac{{\mathrm{y}}^2}{{\mathrm{b}}^2}=1$ then eccentric angle $\mathrm{\theta }$ is

Equation of  of tangent  to the  given ellipse is$\frac{\mathrm{x}}{\mathrm{a}}\mathrm{cos\theta }+\frac{\mathrm{y}}{\mathrm{b}}\mathrm{sin\theta }=1$

 $$\begin{gathered} Given \tan gent is \sqrt{3} \mathrm{bx}+\mathrm{ay}=2\mathrm{ab} \\ \frac{\sqrt{3}}{2} \frac{\mathrm{x}}{\mathrm{a}}+\frac{\mathrm{y}}{\mathrm{b}}.\frac{1}{2}=1 \\ By comparision we get \\ \mathrm{cos\theta }=\frac{\sqrt{3}}{2} , \mathrm{sin\theta }=\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right. \\ \Rightarrow \mathrm{\theta }=\raisebox{1ex}{$\mathrm{\pi }$}\!\left/ \!\raisebox{-1ex}{$6$}\right. \end{gathered}$$