The value of D is....For a fixed constant , let be the locus of a point such that the product of the distance of from and the distance of [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 $D$ is ______.

Smart Achievers · Accepted Answer

(77.14)

$Perpendicular bisector of R S$

$T \equiv (\frac{x_{1} + x_{2}}{2}, \frac{y_{1} + y_{2}}{2})$

$Since, x_{1} + x_{2} = 0 \Rightarrow y_{1} + y_{2} = 2$

$So, T = (0, 1)$ $[from (i)]$

$Equation of R' S' : (y - 1) = - \frac{1}{2} (x - 0) \Rightarrow x + 2 y = 2$

$Let R' (a_{1}, b_{1}) and S' (a_{2}, b_{2})$

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

$On solving x + 2 y = 2 and | 2 x^{2} - {(y - 1)}^{2} | = 3 λ^{2}, we get$

$\Rightarrow$ $| 8 {(y - 1)}^{2} - {(y - 1)}^{2} |$ $= 3 λ^{2}$

$\Rightarrow {(y - 1)}^{2} = {(\frac{\sqrt{3} λ}{\sqrt{7}})}^{2}$

$\Rightarrow y - 1 = \pm \frac{\sqrt{3} λ}{\sqrt{7}}$

$\Rightarrow b_{1} = 1 + \frac{\sqrt{3} λ}{\sqrt{7}}, b_{2} = 1 - \frac{\sqrt{3} λ}{\sqrt{7}}$

$D = 5 {(\frac{2 \sqrt{3} λ}{\sqrt{7}})}^{2} = \frac{5 \times 4 \times 3 λ^{2}}{7} = \frac{5 \times 4 \times 27}{7} = 77.14$

Solution

Want Us to Email you About Special Offers & Updates?