JEE Advanced 2021 Paper 1 · Q09 · Straight Lines

Question

The value of $\lambda^2$ is _____ .

SolveKar AI · Accepted Answer

Locus $C$: $\dfrac{|x\sqrt{2}+y-1|\cdot|x\sqrt{2}-y+1|}{3}=\lambda^2$, i.e., $|2x^2-(y-1)^2|=3\lambda^2$. For intersection with $y=2x+1$: $2x^2-(2x)^2=\pm 3\lambda^2\Rightarrow -2x^2=\pm 3\lambda^2$, giving $x^2=\dfrac{3\lambda^2}{2}$, so $x=\pm\lambda\sqrt{3/2}$. The chord length on line $y=2x+1$ (slope 2, $\sec	heta=\sqrt{5}$): $|RS|=2|\lambda|\sqrt{3/2}\cdot\sqrt{5}=\lambda\sqrt{30}$. Setting $|RS|=\sqrt{270}\Rightarrow \lambda\sqrt{30}=\sqrt{270}\Rightarrow \lambda^2=9$.

JEE Advanced 2021 Paper 1 · Q09 · Straight Lines

Solution