Question

Consider the lines $L_1$ and $L_2$ defined by
$L_1:x\sqrt{2}+y-1=0$ and $L_2:x\sqrt{2}-y+1=0$
For a fixed constant $\lambda ,$ let $C$ be the locus of a point $P$ such that the product of the distance of $P$ from $L_1$ and the distance of $P$ from $L_2$ is ${\lambda }^2$. The line $y=2x+1$ meets $C$ at two points $R$ and $S,$ where the distance between $R$ and $S$ is $\sqrt{270}.$
The value of ${\lambda }^2$ is

Answer 1

Locus $C,$
$\left|\frac{x\sqrt{2}+y-1}{\sqrt{3}}\right|\left|\frac{x\sqrt{2}-y+1}{\sqrt{3}}\right|={\lambda }^2$
$\Rightarrow |2x^2-(y-1{)}^2|=3{\lambda }^2$
$C$ cuts $y-1=2x$ at $R(x_1,y_1)$ and $S(x_2,y_2).$
So,
$|2x^2-(2x{)}^2|=3{\lambda }^2$
$\Rightarrow x=\pm \lambda \sqrt{\frac{3}{2}}$
$\therefore R(\lambda \sqrt{\frac{3}{2}},1+\lambda \sqrt{6})$ and $S(-\lambda \sqrt{\frac{3}{2}},1-\lambda \sqrt{6})$
Distance between $R$ and $S$ is $\sqrt{270}$
$\Rightarrow \sqrt{(\lambda \sqrt{6}{)}^2+(2\lambda \sqrt{6}{)}^2}=\sqrt{270}$
$\therefore {\lambda }^2=9$