Consider the lines L₁ and L₂ defined by L₁: x 2 +y-1=0 and L₂: x 2 -y+1=0 For a fixed constant , let C be the locus of a point P such that the product of the distance of P from L₁ and the distance of P from L₂ is 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 R S meet C at two distinct points R^ and S^ . Let D be the square of the dis

Step-by-step solution

Let the point P is h , k Distance of P from L 1 = h 2 + k - 1 ( 2 ) 2 + 1 2 = h 2 + k - 1 3 Distance of P from L 2 = h 2 - k + 1 ( 2 ) 2 + 1 2 = h 2 - k + 1 3 ∴ The equation of the locus of P is h 2 + k - 1 3 × h 2 - k + 1 3 = λ 2 h 2 + k - 1 3 h 2 - k + 1 3 = λ 2 ⇒ 2 h 2 - ( k - 1 ) 2 = 3 λ 2 Hence, the equation of the locus is 2 x 2 - ( y - 1 ) 2 = 3 λ 2 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 -