Mathematics - Hyperbola Question with Solution | TestHub

$\frac{x^2}{a^2}-\frac{y^2}{b^2}=1....(1)$
$\begin{array}{rl}& \mathrm{P}(4,2\sqrt{3})\\ & {\mathrm{PS}}_1\cdot {\mathrm{PS}}_2=32\\ & \left|{\mathrm{PS}}_1-{\mathrm{PS}}_2\right|=2\mathrm{a}\\ & \mathrm{P}(4,2\sqrt{3})\text{ lies on }\mathrm{H}\\ & \therefore \frac{16}{{\mathrm{a}}^2}-\frac{12}{{\mathrm{b}}^2}=1\end{array}$
$16b^2-12a^2=a^2{b}^2....(2)$
$\begin{array}{rl}& {\left|{\mathrm{PS}}_1-{\mathrm{PS}}_2\right|}^2=4{\mathrm{a}}^2\\ & {\mathrm{PS}}_1^2+{\mathrm{PS}}_2{}^2-2{\mathrm{PS}}_1\cdot {\mathrm{PS}}_2=4{\mathrm{a}}^2\\ & (\mathrm{ae}-4{)}^2+12+(\mathrm{ae}+4{)}^2+12-64=4{\mathrm{a}}^2\\ & 2{\mathrm{a}}^2{\mathrm{e}}^2-8=4{\mathrm{a}}^2\\ & {\mathrm{a}}^2+{\mathrm{b}}^2-4=2{\mathrm{a}}^2\\ & {\mathrm{b}}^2-{\mathrm{a}}^2=4\end{array}$
$\begin{array}{rl}& (2)\&(3)\Rightarrow 16\left(a^2+4\right)-12a^2=a^2\left(a^2+4\right)\\ & \Rightarrow 16a^2+64-12a^2=a^4+4a^2\\ & \Rightarrow a^4=64\\ & \Rightarrow a^2=8\end{array}$
$\begin{array}{rl}& \therefore {\mathrm{b}}^2=12\\ & {\mathrm{p}}^2+{\mathrm{q}}^2=4{\mathrm{b}}^2+\frac{4{\mathrm{b}}^4}{{\mathrm{a}}^2}\\ & =120\end{array}$