Let the lengths of the transverse and conjugate axes of a hyperbola...

Equation of hyperbola is (\frac{\mathrm{x}^2}{\mathrm{a}^2}-\frac{\mathrm{y}^2}{\mathrm{~b}^2}=1) Directrix: (\mathrm{x}=\frac{-9}{5}) and corresponding foci $(-5,0)$ (\Rightarrow-\frac{\mathrm{a}}{\mathrm{e}}=-\frac{9}{5}) and (-\mathrm{ae}=-5) (\Rightarrow \frac{9 e^2}{5}=5 \Rightarrow e=\sqrt{\frac{25}{9}}=\frac{5}{3} \Rightarrow a=3) (\therefore \mathrm{b}^2=\mathrm{a}^2\left(\mathrm{e}^2-1\right)=9\left(\frac{25}{9}-1\right)=16) Hyperbola (\frac{\mathrm{x}^2}{9}-\frac{\mathrm{y}^2}{16}=1) ((\alpha, 2 \sqrt{5})) lie on it (\Rightarrow \frac{\alpha^2}{9}-\frac{20}{16}=1 \Rightarrow \alpha^2=\frac{36}{16} \times 9=\frac{81}{4}) Product for distance of (\left(\mathrm{x}_1 \mathrm{y}_1\right)) from the two foci $$\begin{aligned} & =\left(e x_1+a\right)\left|e x_1-a\right| \\ & =e^2 x_1^2-a^2 \end{aligned}$$)$For$((\alpha, 2 \sqrt{5}) \Rightarrow \mathrm{P}=\frac{25}{9} \cdot \frac{81}{4}-9=\frac{189}{4}) (4 \mathrm{P}=189)$