Mathematics - Hyperbola Question with Solution | TestHub

$\begin{array}{rl}& \mathrm{H}:\frac{{\mathrm{y}}^2}{{\mathrm{b}}^2}-\frac{{\mathrm{x}}^2}{{\mathrm{a}}^2}=1,\mathrm{e}=\sqrt{3}\\ & \mathrm{e}=\sqrt{1+\frac{{\mathrm{a}}^2}{{\mathrm{b}}^2}}=\sqrt{3}\Rightarrow \frac{{\mathrm{a}}^2}{{\mathrm{b}}^2}=2\\ & {\mathrm{a}}^2=2{\mathrm{b}}^2\\ & \text{ length of L.R. }=\frac{2{\mathrm{a}}^2}{\mathrm{b}}=4\sqrt{3}\\ & a=\sqrt{6}\end{array}$ $P(\alpha ,6)$ lie on $\frac{y^2}{3}-\frac{x^2}{6}=1$ $\begin{array}{rl}& 12-\frac{{\alpha }^2}{6}=1\Rightarrow {\alpha }^2=66\\ & \text{ Foci }=(0,\pm \text{ be })=(0,3)\&(0,-3)\end{array}$ Let $d_1\&d_2$ be focal distances of $\mathrm{P}(\alpha ,6)$ $\begin{array}{rl}& {\mathrm{d}}_1=\sqrt{{\alpha }^2+(6+be{)}^2},{\mathrm{d}}_2=\sqrt{{\alpha }^2+(6-be{)}^2}\\ & {\mathrm{d}}_1=\sqrt{66+81},{\mathrm{d}}_2=\sqrt{66+9}\\ & \beta ={\mathrm{d}}_1{\mathrm{d}}_2=\sqrt{147\times 75}=105\\ & {\alpha }^2+\beta =66+105=171\end{array}$
SECTION - B