Equation of normal to $9{\mathrm{x}}^2-25{\mathrm{y}}^2=225$ at $\mathrm{\theta }=\mathrm{\pi }/4$ is
 

Given hyperbola $\frac{{\mathrm{x}}^2}{25}-\frac{{\mathrm{y}}^2}{9}=1$

$\Rightarrow {\mathrm{a}}^2=25, {\mathrm{b}}^2=9 \mathrm{and} given \mathrm{\theta }=\frac{\mathrm{\pi }}{4}$

Now Equation of normal at $\text{'}\mathrm{\theta }\text{'}$ $\frac{\mathrm{ax}}{\mathrm{sec\theta }}+\frac{\mathrm{by}}{\mathrm{tan\theta }}={\mathrm{a}}^2+{\mathrm{b}}^2$

$$\begin{gathered} \Rightarrow \frac{5\mathrm{x}}{\sqrt{2}}+\frac{3\mathrm{y}}{1}=34 \\ \Rightarrow 5\mathrm{x}+3\sqrt{2} \mathrm{y}=34\sqrt{2}. \end{gathered}$$