JEE Advanced 2021 — Paper 1 — Question 46

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