Let the perpendicular bisector of R S meet C at two distinct points R and S. Let D be the square of the distance between R and S. Q. The value of 2 is [2021]

Question

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 $λ,$ 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 $L_{2}$ is $λ_{2}$ . 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' and S'. Let D be the square of the distance between R' and S'. [2021]

Q. The value of $λ^{2}$ is _______.

Smart Achievers · Accepted Answer

(9)

Let locus point $P (x, y) .$

$∴$ $According to question,$ $| \frac{\sqrt{2} x + y - 1}{\sqrt{3}} |$ $| \frac{\sqrt{2} x - y + 1}{\sqrt{3}} |$ $C$

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

$R$ $| 2 x^{2} - {(y - 1)}^{2} |$ $= 3 λ^{2}$

$Let the line y = 2 x + 1 meets C at two points R (x_{1}, y_{1}) and S (x_{2}, y_{2})$

$\Rightarrow y_{1} = 2 x_{1} + 1, y_{2} = 2 x_{2} + 1 \dots (i)$

$\Rightarrow (y_{1} - y_{2}) = 2 (x_{1} - x_{2})$

$∴$ $R S = \sqrt{{(x_{1} - x_{2})}^{2} + {(y_{1} - y_{2})}^{2}}$

$R S = \sqrt{5 {(x_{1} - x_{2})}^{2}} = \sqrt{5}$ $| x_{1} - x_{2} |$

$On solving equations curve C and line y = 2 x + 1, we get$

$| 2 x^{2} - {(2 x)}^{2} |$ $= 3 λ^{2} \Rightarrow x^{2} = \frac{3 λ^{2}}{2}$

$∴$ $R S = \sqrt{5}$ $| \frac{2 \sqrt{3} λ}{\sqrt{2}} |$ $= \sqrt{30} λ = \sqrt{270}$

$\Rightarrow 30 λ^{2} = 270 \Rightarrow λ^{2} = 9$

Solution

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