If the product of the length of the perpendicular from any point on the hyperbola $16{\mathrm{x}}^2 – 25{\mathrm{y}}^2 = 400$ to its asymptotes is P and the angle between the two asymptotes is R then $\mathrm{P} \mathrm{Tan}\frac{\mathrm{\theta }}{2}=$

$$\begin{gathered} \mathrm{Given} \mathrm{hyperbola} \frac{{\mathrm{x}}^2}{25}-\frac{{\mathrm{y}}^2}{16}=1 \\ \Rightarrow {\mathrm{a}}^2=25, {\mathrm{b}}^2=16 \\ \mathrm{Now} \mathrm{P}=\frac{{\mathrm{a}}^2{\mathrm{b}}^2}{{\mathrm{a}}^2+{\mathrm{b}}^2} \\ \Rightarrow \mathrm{P}=\frac{25\times 16}{41}=\frac{400}{41} \\ \mathrm{and} \mathrm{e}=\frac{\sqrt{25+16}}{25}=\frac{\sqrt{41}}{5} \\ \mathrm{Now} \mathrm{\theta }=2{\sec }^{-1}\left(\mathrm{e}\right) \\ \Rightarrow \mathrm{e}=\sec \left(\frac{\mathrm{\theta }}{2}\right) \\ \Rightarrow \sec \left(\frac{\mathrm{\theta }}{2}\right)=\frac{\sqrt{41}}{5} \\ \Rightarrow \tan \frac{\mathrm{\theta }}{2}=\sqrt{{\sec }^2\frac{\mathrm{\theta }}{2}-1}=\frac{4}{5} \\ \mathrm{Now} \mathrm{P} \tan \frac{\mathrm{\theta }}{2}=\frac{400}{41}\times \frac{4}{5}=\frac{320}{41} \end{gathered}$$