Let ‘p’ be the perpendicular distance from the centre C of the hyperbola $\frac{{\mathrm{x}}^2}{{\mathrm{a}}^2}-\frac{{\mathrm{y}}^2}{{\mathrm{b}}^2}=1$ to the tangent drawn at a point R on the hyperbola. If S and S' are the two foci of the hyperbola, then ${\left(\mathrm{RS}+{\mathrm{RS}}^1\right)}^2={\mathrm{\lambda a}}^2\left(1+\frac{{\mathrm{b}}^2}{{\mathrm{p}}^2}\right)$ where $\mathrm{\lambda }$ is

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

Let $\mathrm{R}(\mathrm{a} \sec \mathrm{\theta }, \mathrm{b} \tan \mathrm{\theta })$ be a point tangent at R is

$\begin{array}{l}\frac{\mathrm{x}}{\mathrm{a}}\sec \mathrm{\theta }-\frac{\mathrm{y}}{\mathrm{b}}\tan \mathrm{\theta }-1=0;\mathrm{S}\text{ and }{\mathrm{S}}^1=(\pm \mathrm{ae},0)\\ \mathrm{p}=\frac{\left|1\right|}{\sqrt{\frac{{\sec }^2\mathrm{\theta }}{{\mathrm{a}}^2}+\frac{{\tan }^2\mathrm{\theta }}{{\mathrm{b}}^2}}}\Rightarrow \frac{1}{{\mathrm{p}}^2}=\frac{{\sec }^2\mathrm{\theta }}{{\mathrm{a}}^2}+\frac{{\tan }^2\mathrm{\theta }}{{\mathrm{b}}^2}\\ \Rightarrow \mathrm{RS}=\mathrm{aesec}\mathrm{\theta }-\mathrm{a};{\mathrm{RS}}^1=\mathrm{aesec}\mathrm{\theta }+\mathrm{a}\\ \therefore {\left(\mathrm{RS}+{\mathrm{RS}}^1\right)}^2={\mathrm{\lambda a}}^2\left(1+\frac{{\mathrm{b}}^2}{{\mathrm{p}}^2}\right)\\ {\left(\mathrm{RS}+{\mathrm{RS}}^1\right)}^2={\mathrm{\lambda a}}^2(1+\frac{b^{2}se{c}^2{\theta }^{ }}{a^2}+{\tan }^2\theta )\\ \Rightarrow 4{\mathrm{a}}^2\cdot \left(\frac{{\mathrm{a}}^2+{\mathrm{b}}^2}{{\mathrm{a}}^2}\right){\sec }^2\mathrm{\theta }=\mathrm{\lambda }\left({\mathrm{a}}^2+{\mathrm{b}}^2\right){\sec }^2\mathrm{\theta }\\ \therefore \mathrm{\lambda }=4\end{array}$