Consider the lines $L_{1}$ and $L_{2}$ 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 $L_{1}$ 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'. The value of $\lambda^{2}$ is ____.

JEE Advanced · 2021 · 03 Oct 2021 · Paper 1 · Straight Lines

Consider the lines

L_{1}

and

L_{2}

defined by

L_{1} : x 2 + y - 1 = 0

and

L_{2} : x 2 - y + 1 = 0

For a fixed constant

λ

, 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}

λ^{2}

. The line

y = 2 x + 1

meets C at two points R and S, where the distance between R and S is

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

λ^{2}

is ____.

Answer: 9

For point P(x,y), the condition gives

\frac{2 x + y - 1}{3} \frac{2 x - y + 1}{3} = λ^{2}

, which simplifies to

\frac{2 x ^{2} - ( y - 1 ) ^{2}}{3} = λ^{2}

, so C:

∣2 x^{2} - (y - 1)^{2} ∣ = 3 λ^{2}

. Substituting

y = 2 x + 1

into the curve equation:

∣2 x^{2} - (2 x)^{2} ∣ = 3 λ^{2}

gives

∣ - 2 x^{2} ∣ = 3 λ^{2}

, so

x^{2} = \frac{3 λ ^{2}}{2}

. For points R

(x_{1}, y_{1})

and S

(x_{2}, y_{2})

on the line,

R S = 5 ∣ x_{1} - x_{2} ∣ = 5 \cdot \frac{2 3 λ}{2} = 30 λ^{2} = 270

. Therefore,

30 λ^{2} = 270

, giving

λ^{2} = 9

JEE Advanced · 2021 · 03 Oct 2021 · Paper 1 · Numerical answer