JEE Advanced 2021 Paper 1 · Q10 · Straight Lines

Correct answer:77.14

Solution

Midpoint of $RS$ on $y=2x+1$: at $x=0$, midpoint $=(0,1)$. Perpendicular bisector: $y-1=-\dfrac{1}{2}x$, i.e., $y=-\dfrac{x}{2}+1$. Substituting into $|2x^2-(y-1)^2|=3\lambda^2=27$: $\left|2x^2-\dfrac{x^2}{4}\right|=27\Rightarrow \dfrac{7x^2}{4}=27\Rightarrow x^2=\dfrac{108}{7}$, so $x=\pm\sqrt{108/7}$. Length $|R'S'|=2\sqrt{108/7}\cdot\sec\phi$ where slope $-1/2$ gives $\sec\phi=\sqrt{5}/2\cdot 2=\sqrt{5/4+1}=\sqrt{5}/2$. Computing $D=|R'S'|^2=\dfrac{4\cdot 108}{7}\cdot\dfrac{5}{4}=\dfrac{540}{7}=77.14$.

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