Suppose the hyperbola $\frac{{\mathrm{x}}^2}{{\mathrm{a}}^2}-\frac{{\mathrm{y}}^2}{{\mathrm{b}}^2}=1,$ where $\mathrm{a}>\mathrm{b}$, has a tangent of slope $\sqrt{2}$ ,then  eccentricity  lies in the interval

$$\begin{gathered} \mathrm{Given} \mathrm{hyperbola} \frac{{\mathrm{x}}^2}{{\mathrm{a}}^2}-\frac{{\mathrm{y}}^2}{{\mathrm{b}}^2}=1 \\ \mathrm{let} \mathrm{P}=(\mathrm{a} \mathrm{sec\theta }, \mathrm{b} \mathrm{tan\theta }) \mathrm{be} \mathrm{any} \mathrm{point} \mathrm{on} \mathrm{hyperbola} \\ \mathrm{diff} \mathrm{w}.\mathrm{r}. \mathrm{to} \text{'}\mathrm{x}\text{'} \mathrm{on} \mathrm{b}.\mathrm{s} \\ \frac{2\mathrm{x}}{{\mathrm{a}}^2}-\frac{2\mathrm{y}{\displaystyle \frac{\mathrm{dy}}{\mathrm{dx}}}}{{\mathrm{b}}^2}=0 \\ \Rightarrow \frac{2\mathrm{x}}{{\mathrm{a}}^2}=\frac{2\mathrm{y}}{{\mathrm{b}}^2} \frac{\mathrm{dy}}{\mathrm{dx}} \\ \Rightarrow \frac{\mathrm{dy}}{\mathrm{dx}}=\frac{{\mathrm{b}}^2}{{\mathrm{a}}^2}\left(\frac{\mathrm{x}}{\mathrm{y}}\right) \\ Now \mathrm{slope} \mathrm{of} \mathrm{tangent} \mathrm{at} \mathrm{P}=\left(\frac{{\mathrm{b}}^2}{{\mathrm{a}}^2}\right) \left(\frac{\mathrm{a} \mathrm{sec\theta }}{\mathrm{b} \mathrm{tan\theta }}\right) \\ \Rightarrow 2\sqrt{2}=\frac{\mathrm{b}}{\mathrm{a}}\frac{\mathrm{sec\theta }}{\mathrm{tan\theta }}\Rightarrow \frac{\mathrm{b}}{\mathrm{a}}=2\sqrt{2} \mathrm{sin\theta } \\ \mathrm{Now} \mathrm{e}=\sqrt{1+\frac{{\mathrm{b}}^2}{{\mathrm{a}}^2}} \\ =\sqrt{1+8{\sin }^2\mathrm{\theta }} \\ \mathrm{for} \mathrm{hyperbola} \mathrm{e}>1, so 0<{\sin }^2\theta \le 1 \\ \Rightarrow \mathrm{e}\in (1, 3] \end{gathered}$$