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Points and Straight Lines - JEE Mathematics question (JEE Advanced 2021 Paper 1)

Comprehension · Questions 1–1

Read the passage given below and answer the questions that follow.

Consider the lines $L_1$ and $L_2$ defined by $$L_1: x\sqrt{2}+y-1=0 \text{ and } L_2: x\sqrt{2}-y+1=0$$ 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 $R'$ and $S'$. Let $D$ be the square of the distance between $R'$ and $S'$.
JEE Advanced 2021 Paper 1Question 1
The value of $\lambda^2$ is ___ .
Points and Straight Lines

Answer$9$

Solution

Product of distances $=\frac{|2x^2-(y-1)^2|}{3}=\lambda^2$. On $y=2x+1$: $2x^2=3\lambda^2$. $RS^2=5(2x)^2=20x^2=270$ gives $x^2=\frac{27}{2}$, so $\lambda^2=\frac{2x^2}{3}=9$.

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