Mathematics - Hyperbola Question with Solution | TestHub

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 $-\mathrm{ae}=-5$ $\Rightarrow \frac{9e^2}{5}=5\Rightarrow e=\sqrt{\frac{25}{9}}=\frac{5}{3}\Rightarrow a=3$ $\therefore {\mathrm{b}}^2={\mathrm{a}}^2\left({\mathrm{e}}^2-1\right)=9\left(\frac{25}{9}-1\right)=16$ 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 $\begin{array}{rl}& =\left(e{x}_1+a\right)\left|e{x}_1-a\right|\\ & =e^2{x}_1^2-a^2\end{array}$ For $(\alpha ,2\sqrt{5})\Rightarrow \mathrm{P}=\frac{25}{9}\cdot \frac{81}{4}-9=\frac{189}{4}$ $4\mathrm{P}=189$