The tangent at a 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, –b) and the normal at P passes through the point $(2\mathrm{a}\sqrt{2},0)$ then eccentricity of the hyperbola is

$Given hyperbola \frac{x^2}{a^2}-\frac{y^2}{b^2}=1$

Equation of tangent at $\mathrm{P}\left({\mathrm{x}}_1,{\mathrm{y}}_1\right)$ is $\frac{{\mathrm{xx}}_1}{{\mathrm{a}}^2}-\frac{{\mathrm{yy}}_1}{{\mathrm{b}}^2}=1$ …..(1) $\Rightarrow$ (1) passing through (0, -b) $\Rightarrow {\mathrm{y}}_1=\mathrm{b}$

$Equation of normal at (x_1,y_1) is \frac{a^{2}x}{x_1}+\frac{b^{2}y}{y_1}=a^2+b^2$

Equation of normal passing through $(2\sqrt{2}\mathrm{a},0)$ then

$$\begin{gathered} \frac{a^2\left(2\sqrt{2}a\right)}{x_1}+0=a^2{e}^2 \\ \Rightarrow x_1=\frac{2\sqrt{2}a}{e^2} \\ Since the point (x_1,y_1) lies on hyperbola then \\ \frac{8a^2}{a^2{e}^4}-1=1 \\ \Rightarrow \frac{8}{e^4}=2 \\ \Rightarrow e^{4_{ }}=4 \\ \Rightarrow e=\sqrt{2} \end{gathered}$$