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 D 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 D is ____.

Answer: 77.14

Solution

From the previous part,

λ^{2} = 9

. The midpoint T of RS is at

(\frac{x _{1} + x _{2}}{2}, \frac{y _{1} + y _{2}}{2})

. Since

x_{1} + x_{2} = 0

, we have T=(0,1). The perpendicular bisector of RS has slope

- 1/2

(perpendicular to slope 2), so its equation is

y - 1 = - \frac{1}{2} (x - 0)

, or

x + 2 y = 2

. For points R'

(a_{1}, b_{1})

and S'

(a_{2}, b_{2})

on this line and the curve,

D = (a_{1} - a_{2})^{2} + (b_{1} - b_{2})^{2} = 5 (b_{1} - b_{2})^{2}

. Substituting

x = 2 - 2 y

into

∣2 x^{2} - (y - 1)^{2} ∣ = 27

∣8 (y - 1)^{2} - (y - 1)^{2} ∣ = 27

gives

(y - 1)^{2} = \frac{27}{7}

, so

y - 1 = \pm \frac{27}{7} = \pm \frac{3 3}{7}

. Therefore,

D = 5 (\frac{6 3}{7})^{2} = 5 \cdot \frac{108}{7} = \frac{540}{7} \approx 77.14

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

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

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