Consider the lines $L_{1}$ and $L_{2}$ defined by $ L_{1}: x \sqrt{2}+y-1=0 ext { 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 $\lambda^{2}$.…

Question

Consider the lines $L_{1}$ and $L_{2}$ defined by
$
L_{1}: x \sqrt{2}+y-1=0 \text { 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 $y=2 x+1$ 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 $R^{\prime}$ and $S^{\prime}$. Let $D$ be the square of the distance between $R^{\prime}$ and $S^{\prime}$.

The value of $D$ is

Aicharya · Accepted Answer

From the first question

The equation of the locus is $|2 x^{2} - (y - 1)^{2}| = 27$

The line is $y = 2 x + 1$ or $y - 1 = 2 x$

By substituting the value of $y$ in the equation of the curve $C$ , we get

$|2 x^{2} - (y - 1)^{2}| = 27$

$\Rightarrow |2 x^{2} - (2 x)^{2}| = 27$

$\Rightarrow 2 x^{2} = 27$

$\Rightarrow x = \pm 3 \sqrt{\frac{3}{2}}$

$\Rightarrow x_{1}, x_{2} = \pm 3 \sqrt{\frac{3}{2}}$

Let $M$ be the mid-point of $R'$ and $S'$

So, the $x$ coordinate of $T$ is $\frac{x_{1} + x_{2}}{2} = 0$

It lies on $y = 2 x + 1$

So, the coordinates of $T$ are $(0, 1)$

Slope of the line perpendicular to $y = 2 x + 1$ is $\frac{- 1}{2}$

So, the equation of perpendicular bisector is

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

Or, $x + 2 y = 2$

Let coordinate of $R'$ and $S'$ are $(p_1, q_1)$ and $(p_2, q_2)$, we get

$D = {(p_{2} - p_{1})}^{2} + {(q_{2} - q_{1})}^{2}$

Since both the points satisfy the equation of the line, we get

$D = {[2 (q_{2} - q_{1})]}^{2} + {(q_{2} - q_{1})}^{2}$

$D = 5 {(q_{2} - q_{1})}^{2}$

Solving, $x + 2 y = 2$ with $2 x^{2} - (y - 1)^{2} = 27$ , we get

$2 {(2 - 2 y)}^{2} - (y - 1)^{2} = 27$

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

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

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

$\Rightarrow q_{1}, q_{2} = 1 \pm 3 \sqrt{\frac{3}{7}}$

So, ${(q_{2} - q_{1})}^{2} = {(6 \sqrt{\frac{3}{7}})}^{2}$

Hence, $D = 5 {(q_{2} - q_{1})}^{2} = 5 \times {(6 \sqrt{\frac{3}{7}})}^{2} = 77.14$

Consider the lines $L_{1}$ and $L_{2}$ defined by $ L_{1}: x \sqrt{2}+y-1=0 \text { and } L_{2}: x…

Solution