Mathematics - Hyperbola Question with Solution | TestHub

Equation of normal to hyperbola at $\left(x_1,y_1\right)$ is $$\begin{array}{c}\frac{a^{2}x}{x_1}+\frac{b^{2}y}{y_1}=a^2+b^2\\ \therefore \text{ At }(6,3),\frac{a^{2}x}{6}+\frac{b^{2}y}{3}=a^2+b^2\end{array}$$ It passes throught $(9,0)$. $$\begin{array}{rl}& \text{ Now, }\frac{a^2\cdot 9}{6}=a^2+b^2\\ & \Rightarrow \frac{3a^2}{2}-a^2=b^2\Rightarrow \frac{a^2}{b^2}=2\\ & \therefore e^2=1+\frac{b^2}{a^2}=1+\frac{1}{2}\Rightarrow e=\sqrt{\frac{3}{2}}\end{array}$$