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 $\sqrt{270}$.
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

Answer: 9

Locus $C=\left|\frac{(x\sqrt{2}+y-1)(x\sqrt{2}-y+1)}{\sqrt{3}}\right|={\lambda }^2$
$2x^2-(y-1{)}^2=\pm 3{\lambda }^2$
for intersection with $y=2x+1$
$2x^2-(2x{)}^2=\pm 3{\lambda }^2$
$-2x^2=-3{\lambda }^2$ (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, $\sqrt{30}|\lambda |=\sqrt{270}(\lambda =\pm 3)$
${\lambda }^2=9$