The ordinate of any point P on the hyperbola $25{\mathrm{x}}^2 – 16{\mathrm{y}}^2 = 400$ is produced to cut the asymptotes in the points Q and R,then the value of QP.PR is equal to$\mathrm{\_\_\_\_\_}$

$$\begin{gathered} \mathrm{Given} \mathrm{hyperbola} \frac{{\mathrm{x}}^2}{16}-\frac{{\mathrm{y}}^2}{25}=1 \\ where {\mathrm{a}}^2=16, {\mathrm{b}}^2=25 \\ \mathrm{Asymptotes} \mathrm{of} \mathrm{hyperbola} \mathrm{are} \mathrm{y}=\pm \frac{5}{4}\mathrm{x} \\ Let \mathrm{P}=\left({\mathrm{x}}_1, \frac{5}{4}\sqrt{{{\mathrm{x}}_1}^2-16}\right) \mathrm{be} \mathrm{any} \mathrm{point} \mathrm{on} \mathrm{hyperbola}. \\ \mathrm{Now} \mathrm{Q}=\left({\mathrm{x}}_1, \frac{5}{4} {\mathrm{x}}_1\right), \mathrm{R}=\left({\mathrm{x}}_1, \frac{-5}{4}{\mathrm{x}}_1\right) \mathrm{are} \mathrm{the} \mathrm{points} \mathrm{where} \mathrm{ordinate} \mathrm{meets} \mathrm{the} \mathrm{asymptotes}. \\ \mathrm{Now} \mathrm{QP}=\left(\frac{5}{4}{\mathrm{x}}_1-\frac{5}{4}\sqrt{{{\mathrm{x}}_1}^2-16}\right) \\ \mathrm{PR}=\left(\frac{5}{4}\sqrt{{{\mathrm{x}}_1}^2-16}+\frac{5}{4}{\mathrm{x}}_1\right) \\ \mathrm{Now} \mathrm{QP}.\mathrm{PR}=\left(\frac{5}{4}{\mathrm{x}}_1-\frac{5}{4}\sqrt{{{\mathrm{x}}_1}^2-16}\right) \left(\frac{5}{4}{\mathrm{x}}_1+\frac{5}{4}\sqrt{{{\mathrm{x}}_1}^2-16}\right) \\ = \frac{25{{\mathrm{x}}_1}^2}{16} -\frac{25}{16} ({{\mathrm{x}}_1}^2-16) \\ =\frac{25}{16}\times 16=25 \end{gathered}$$