QJEEAdv21Maae9b: JEE Advanced 2021 Mathematics - Class 11 JEEAdv21Maae9b Solution: Consider the lines L 1 and L 2 defined by and For a fixed…

Question

Verified Solution • 2025–26 Session moderate

Consider the lines L₁ and L₂ defined by

{L_1}:x\sqrt 2 + y - 1 = 0

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 L₁ and the distance of P from L₂ 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'.

The value of

\lambda

² is __________.

Accepted Answer

According to the question,

C:\left| {{{x\sqrt 2 + y - 1} \over {\sqrt 3 }}} \right|\left| {{{x\sqrt 2 - y + 1} \over {\sqrt 3 }}} \right| = {\lambda ^2}

\Rightarrow C:{{\left| {{{(x\sqrt 2 )}^2} - {{(y - 1)}^2}} \right|} \over {\sqrt 3 \times \sqrt 3 }} = {\lambda ^2}

\Rightarrow C:\left| {2{x^2} - {{(y - 1)}^2}} \right| = 3{\lambda ^2}

Let R

\equiv

(x₁, y₁) and S(x₂, y₂)

\because

C cuts y

-

1 = 2x at R and S.

So,

\left| {2{x^2} - 4{x^2}} \right| = 3{\lambda ^2}

\Rightarrow x = \pm \sqrt {{3 \over 2}} \left| \lambda \right|

\therefore

\left| {{x_1} - {x_2}} \right| = \sqrt 6 \left| \lambda \right|

and

\left| {{y_1} - {y_2}} \right| = 2\left| {{x_1} - {x_2}} \right| = 2\sqrt 6 \left| \lambda \right|

\because

RS² = 270 (given)

\Rightarrow {({x_1} - {x_2})^2} + {({y_1} - {y_2})^2} = 270

\Rightarrow {(\sqrt 6 \lambda )^2} + {(2\sqrt 6 \left| \lambda \right|)^2} = 270

\Rightarrow 30{\lambda ^2} = 270 \Rightarrow {\lambda ^2} = 9

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