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 $D$ is _____

Answer 1

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