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 $y=2 x+1$ meets $C$ at two points $R$ and $S$, where the distance between $R$ and $S$ is $\sqrt{270}$.
Let the perpendicular bisector of $RS$ meet $C$ at two distinct points $R ^{\prime}$ and $S ^{\prime}$. Let $D$ be the square of the distance between $R ^{\prime}$ and $S ^{\prime}$
The value of $\lambda^2$ is _____ .

Answer 1

Locus $C=\left|\frac{(x \sqrt{2}+y-1)(x \sqrt{2}-y+1)}{\sqrt{3}}\right|=\lambda^{2}$
$2 x^{2}-(y-1)^{2}=\pm 3 \lambda^{2}$
for intersection with $y=2 x+1$
$2 x^{2}-(2 x)^{2}=\pm 3 \lambda^{2}$
$-2 x^{2}=-3 \lambda^{2}$ (taking $-$ ve sign )
$x=\pm \sqrt{\frac{3}{2}} \lambda$
Distance between $R$ and $S =2\left|\sqrt{\frac{3}{2}} \lambda\right| \sec \theta (\tan \theta$ is slope of line)
$=\sqrt{6}|\lambda| \sqrt{5}$
So, $\sqrt{30}|\lambda|=\sqrt{270}(\lambda=\pm 3)$
$\lambda^{2}=9$