Let P(6,3) be a point on the hyperbola $\frac{{\mathrm{x}}^2}{{\mathrm{a}}^2}-\frac{{\mathrm{y}}^2}{{\mathrm{b}}^2}=1.$ If the normal at the point P intersects the x-axis at (9,0), then the eccentricity of the hyperbola =

$$\begin{gathered} Given hyperbola \frac{{\mathrm{x}}^2}{{\mathrm{a}}^2}-\frac{{\mathrm{y}}^2}{{\mathrm{b}}^2}=1 \\ slope of \tan gent at p(x_1,y_1)=\frac{b^2{x}_1}{a^2{y}_1} \\ \mathrm{equation} \mathrm{of} \mathrm{normal} \mathrm{at} ({\mathrm{x}}_1, {\mathrm{y}}_1) \mathrm{is} y-y_1=\frac{-a^2{y}_1}{b^2{x}_1}(x-x_1) \\ \mathrm{Now} \mathrm{equation} \mathrm{of} \mathrm{normal} \mathrm{at} \mathrm{P}(6, 3) \mathrm{is} \\ \Rightarrow y-3=\frac{-3a^2}{6b^2}(x-6) \\ \mathrm{Since}, \mathrm{it} \mathrm{intersects} \mathrm{x}-\mathrm{axis} \mathrm{at} (9, 0) \mathrm{then} \\ 0-3=\frac{-3a^2}{6b^2}(9-6) \\ \Rightarrow \frac{3{\mathrm{a}}^2}{6b^2}=1 \\ \Rightarrow {\mathrm{a}}^2=2{\mathrm{b}}^2 \\ \mathrm{Now} \mathrm{e}=\sqrt{\frac{{\mathrm{a}}^2+{\mathrm{b}}^2}{{\mathrm{a}}^2}}=\sqrt{\frac{3{\mathrm{b}}^2}{2{\mathrm{b}}^2}}=\sqrt{\frac{3}{2}} \end{gathered}$$