If any point on a hyperbola is $\left(3\tan \mathrm{\varphi }, 2\sec \mathrm{\varphi }\right)$ then the eccentricity of the hyperbola is

clearly given point $\left(3\tan \cancel{0}, 2\mathrm{ec}\cancel{0}\right)$ satisfies the hyperbola $\frac{{\mathrm{x}}^2}{9}-\frac{{\mathrm{y}}^2}{4}=-1$

$$\begin{gathered} \Rightarrow {\mathrm{a}}^2=9, {\mathrm{b}}^2=4 \\ \mathrm{Now} \mathrm{e}=\sqrt{\frac{{\mathrm{a}}^2+{\mathrm{b}}^2}{{\mathrm{b}}^2}} \\ =\sqrt{\frac{13}{4}}=\sqrt{\frac{13}{2}} \end{gathered}$$