A tangent drawn to hyperbola $\frac{{\mathrm{x}}^2}{{\mathrm{a}}^2}-\frac{{\mathrm{y}}^2}{{\mathrm{b}}^2}=1$ at $\mathrm{P}\left(\frac{\mathrm{\pi }}{6}\right)$ forms a triangle of area $3a^2$square units, with coordinate axes, then the square of its eccentricity is

$$\begin{gathered} Given hyperbola \frac{{\mathrm{x}}^2}{{\mathrm{a}}^2}-\frac{{\mathrm{y}}^2}{{\mathrm{b}}^2}=1 \\ \mathrm{P}\left(\frac{\mathrm{\pi }}{6}\right)=(\mathrm{a} \sec {30}^0, \mathrm{b} \tan {30}^0) \\ =\left(\frac{2\mathrm{a}}{\sqrt{3}}, \frac{\mathrm{b}}{\sqrt{3}}\right) \\ \mathrm{Now} \mathrm{equation} \mathrm{of} \mathrm{tangent} \mathrm{at} \mathrm{P} \mathrm{is} \\ \Rightarrow \frac{\mathrm{x}\left({\displaystyle \frac{2a}{\sqrt{3}}}\right)}{{\mathrm{a}}^2}-\frac{\mathrm{y}\left({\displaystyle \frac{\mathrm{b}}{\sqrt{3}}}\right)}{{\mathrm{b}}^2}=1 \\ \Rightarrow \frac{\mathrm{x}}{{\displaystyle \frac{\sqrt{3} \mathrm{a}}{2}}}-\frac{\mathrm{y}}{\sqrt{3} \mathrm{b}}=1 — ① \\ \mathrm{Given} \mathrm{that} \mathrm{area} \mathrm{of} \mathrm{triangle} \mathrm{formed} by ① \mathrm{with} \mathrm{coord}i\mathrm{nate} \mathrm{axes}=3{\mathrm{a}}^2 \end{gathered}$$

$$\begin{gathered} \Rightarrow \frac{1}{2}\left|\frac{\sqrt{3} \mathrm{a}}{2}\times \left(-\sqrt{3} \mathrm{b}\right)\right|=3{\mathrm{a}}^2 \\ \Rightarrow \mathrm{b}=4\mathrm{a} \\ \Rightarrow {\mathrm{b}}^2=16{\mathrm{a}}^2 \\ \Rightarrow {\mathrm{a}}^2({\mathrm{e}}^2-1)=16{\mathrm{a}}^2 \\ \Rightarrow {\mathrm{e}}^2-1=16 \\ \Rightarrow {\mathrm{e}}^2=17 \end{gathered}$$