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

Answer: 77.14

Equation of perpendicular bisector $y=-\frac{1}{2}x+1$
For point of intersection $2x^2-\frac{1}{4}{x}^2=\pm 3{\lambda }^2$
$x=\pm \sqrt{\frac{12}{7}}\lambda$ (taking + ve sign)
Distance $=\left|2\cdot \sqrt{\frac{12}{7}}\cdot 3\cdot \sec \theta \right|$
$=2\cdot \sqrt{\frac{12}{7}}\cdot 3\cdot \sqrt{\frac{5}{2}}$
$=3\cdot \sqrt{\frac{60}{7}}$
$D=\frac{9\times 60}{7}=77.14$