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JEE Advanced · Mathematics · 9. Straight Lines

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 \(y=2 x+1\) meets \(C\) at two points \(R\) and \(S\), where the distance between \(R\) and \(S\) is \(\sqrt{270}\).
Let the perpendicular bisector of \(R S\) 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

  1. A 77.14
  2. B 75.45
  3. C 60.12
  4. D 50.15
Verified Solution

Answer & Solution

Correct Answer

(A) 77.14

Step-by-step Solution

Detailed explanation

From the first question The equation of the locus is 2x2-(y-1)2=27
From JEE Advanced
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