Let the lengths of the transverse and conjugate axes of a hyperbola in standard form be 2a and 2b, respectively, and one focus and the corresponding…

Equation of hyperbola is $\frac{\mathrm{x}^2}{\mathrm{a}^2}-\frac{\mathrm{y}^2}{\mathrm{~b}^2}=1$a2x2​− b2y2​=1 Directrix: $\mathrm{x}=\frac{-9}{5}$x=5−9​ and corresponding foci $(-5,0)$(−5,0) $\Rightarrow-\frac{\mathrm{a}}{\mathrm{e}}=-\frac{9}{5}$⇒−ea​=−59​ and $-\mathrm{ae}=-5$−ae=−5 $\Rightarrow \frac{9 e^2}{5}=5 \Rightarrow e=\sqrt{\frac{25}{9}}=\frac{5}{3} \Rightarrow a=3$⇒59e2​=5⇒e=925​​=35​⇒a=3 $\therefore \mathrm{b}^2=\mathrm{a}^2\left(\mathrm{e}^2-1\right)=9\left(\frac{25}{9}-1\right)=16$∴b2=a2(e2−1)=9(925​−1)=16 Hyperbola $\frac{\mathrm{x}^2}{9}-\frac{\mathrm{y}^2}{16}=1$9x2​−16y2​=1 $(\alpha, 2 \sqrt{5})$(α,25​) lie on it $\Rightarrow \frac{\alpha^2}{9}-\frac{20}{16}=1 \Rightarrow \alpha^2=\frac{36}{16} \times 9=\frac{81}{4}$⇒9α2​−1620​=1⇒α2=1636​×9=481​ Product for distance of $\left(\mathrm{x}_1 \mathrm{y}_1\right)$(x1​y1​) from the two foci $\begin{aligned} & =\left(e x_1+a\right)\left|e x_1-a\right| \\ & =e^2 x_1^2-a^2 \end{aligned}$​=(ex1​+a)∣ex1​−a∣=e2x12​−a2​ For $(\alpha, 2 \sqrt{5}) \Rightarrow \mathrm{P}=\frac{25}{9} \cdot \frac{81}{4}-9=\frac{189}{4}$(α,25​)⇒P=925​⋅481​−9=4189​ $4 \mathrm{P}=189$4P=189