Prove that the product of the perpendiculars from any point on the hyperbola $\frac{{\mathrm{x}}^2}{{\mathrm{a}}^2}-\frac{{\mathrm{y}}^2}{{\mathrm{b}}^2}=1$ to its asymptotes is constant.

Let P(asec$\mathrm{\theta }$,b tan$\mathrm{\theta }$) be any point on the hyperbola $\frac{{\mathrm{x}}^2}{{\mathrm{a}}^2}-\frac{{\mathrm{y}}^2}{{\mathrm{b}}^2}=1$
The asymptotes are $\frac{{\mathrm{x}}^2}{{\mathrm{a}}^2}-\frac{{\mathrm{y}}^2}{{\mathrm{b}}^2}=0$
Product of perpendicular from  P to the asymptotes
$\begin{array}{l}=\left|\frac{\sec \mathrm{\theta }+\tan \mathrm{\theta }}{\sqrt{\left(\frac{1}{{\mathrm{a}}^2}+\frac{1}{{\mathrm{b}}^2}\right)}}\right|\left|\frac{\sec \mathrm{\theta }-\tan \mathrm{\theta }}{\sqrt{\left(\frac{1}{{\mathrm{a}}^2}+\frac{1}{{\mathrm{b}}^2}\right)}}\right|\\ =\frac{{\sec }^2\mathrm{\theta }-{\tan }^2\mathrm{\theta }}{\left(\frac{1}{{\mathrm{a}}^2}+\frac{1}{{\mathrm{b}}^2}\right)}=\frac{1}{\frac{1}{{\mathrm{a}}^2}+\frac{1}{{\mathrm{b}}^2}}=\frac{{\mathrm{a}}^2{\mathrm{b}}^2}{{\mathrm{a}}^2+{\mathrm{b}}^2}=\text{ constant }\end{array}$