The tangent at point P on the hyperbola $\frac{{\mathrm{x}}^2}{{\mathrm{a}}^2}-\frac{{\mathrm{y}}^2}{{\mathrm{b}}^2}=1$ passes through the point $(0,-\mathrm{b})$ and the normal at point P passes through the point $(2\sqrt{2}\mathrm{a},0)$. If e denote the eccentricity of hyperbola then find the value of e².

The equation of tangent at $\mathrm{P}(\mathrm{x},\mathrm{y})$ is $\frac{{\mathrm{xx}}_1}{{\mathrm{a}}^2}-\frac{{\mathrm{yy}}_1}{{\mathrm{b}}^2}=1$
It passes through $(0,-\mathrm{b})$, so $0+\frac{{\mathrm{y}}_1}{\mathrm{b}}=1\Rightarrow {\mathrm{y}}_1=\mathrm{b}$
Also, equation of normal at $\mathrm{P}({\mathrm{x}}_1,{\mathrm{y}}_1)$ is $\frac{{\mathrm{a}}^2\mathrm{x}}{{\mathrm{x}}_1}+\frac{{\mathrm{b}}^2\mathrm{y}}{{\mathrm{y}}_1}={\mathrm{a}}^2{\mathrm{e}}^2$
It passes through $(2\sqrt{2}\mathrm{a},0)$,so $\frac{{\mathrm{a}}^2\left(2\sqrt{2}\mathrm{a}\right)}{{\mathrm{x}}_1}={\mathrm{a}}^2{\mathrm{e}}^2\Rightarrow {\mathrm{x}}_1=\frac{2\sqrt{2}\mathrm{a}}{{\mathrm{e}}^2}$
So,$\mathrm{P}({\mathrm{x}}_1=\frac{2\sqrt{2}\mathrm{a}}{{\mathrm{e}}^2},{\mathrm{y}}_1=\mathrm{b})$
As $\mathrm{P}({\mathrm{x}}_1,{\mathrm{y}}_1)$lies on the hyperbola $\frac{{\mathrm{x}}^2}{{\mathrm{a}}^2}-\frac{{\mathrm{y}}^2}{{\mathrm{b}}^2}=1,$so $\frac{8{\mathrm{a}}^2}{{\mathrm{e}}^4{\mathrm{a}}^2}-\frac{{\mathrm{b}}^2}{{\mathrm{b}}^2}=1$
$\Rightarrow \frac{8}{{\mathrm{e}}^4}=2\Rightarrow {\mathrm{e}}^4=4\Rightarrow {\mathrm{e}}^2=2$