Given hyperbola $\frac{{\mathrm{x}}^2}{{\mathrm{a}}^2}-\frac{{\mathrm{y}}^2}{{\mathrm{b}}^2}=1$

$$\begin{gathered} Let P\left(\theta \right)=(asce\theta ,b\tan \theta ) be any point on hyperbola \\ Now equation of \tan gent at P\left(\theta \right) is \frac{x}{a}sec\theta -\frac{y}{b}\tan \theta =1 \\ slope of \tan gent=\frac{bsec\theta }{a\tan \theta } \end{gathered}$$
Slope of the normal at P(asec$\mathrm{\theta }$,btan$\mathrm{\theta }$) on the hyperbola $=-\left(\frac{\mathrm{atan}\mathrm{\theta }}{\mathrm{bsec}\mathrm{\theta }}\right)$
the equation of normal at P is
 $\begin{array}{l}\mathrm{y}-\mathrm{btan}\mathrm{\theta }=-\frac{\mathrm{atan}\mathrm{\theta }}{\mathrm{bsec}\mathrm{\theta }}(\mathrm{x}-\mathrm{asec}\mathrm{\theta })\\ \Rightarrow \frac{\mathrm{b}}{\tan \mathrm{\theta }}(\mathrm{y}-\mathrm{btan}\mathrm{\theta })=-\frac{\mathrm{a}}{\sec \mathrm{\theta }}(\mathrm{x}-\mathrm{asec}\mathrm{\theta })\\ \Rightarrow \frac{\mathrm{ax}}{\sec \mathrm{\theta }}+\frac{\mathrm{by}}{\tan \mathrm{\theta }}={\mathrm{a}}^2+{\mathrm{b}}^2\end{array}$