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

Question

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 ​ 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 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

TestHub · 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 $R^{'} & S^{'}$

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 $R^{'} & S^{'}$ are $(p_{1}, q_{1}) & (p_{2}, q_{2})$ , 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, $D = 5 {(q_{2} - q_{1})}^{2} = 5 \times {(6 \sqrt{\frac{3}{7}})}^{2} = 77.14$

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