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 $D$ 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 $\sqrt{270}$
$\Rightarrow \sqrt{(\lambda \sqrt{6}{)}^2+(2\lambda \sqrt{6}{)}^2}=\sqrt{270}$
$\therefore {\lambda }^2=9$

Now, the curve $C$ is
$\Rightarrow |2x^2-(y-1{)}^2|=27\cdots (1)$
Mid-point of $RS$ is (0, 1) and perpendicular bisector is
$y=-\frac{1}{2}x+1\cdots \left(2\right)$
From $\left(1\right)$ and $\left(2\right),$ we have
$\frac{7x^2}{2}=27\Rightarrow x=\pm 6\sqrt{\frac{3}{7}}$
$\therefore R^{\prime }(6\sqrt{\frac{3}{7}},1-3\sqrt{\frac{3}{7}})$ and $S^{\prime }(-6\sqrt{\frac{3}{7}},1+3\sqrt{\frac{3}{7}})$
$D=(R^{\prime }{S}^{\prime }{)}^2={\left(12\sqrt{\frac{3}{7}}\right)}^2+{\left(6\sqrt{\frac{3}{7}}\right)}^2$
$\Rightarrow D=\frac{3}{7}\times 180=\frac{540}{7}$
$\therefore D\approx 77.14$