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

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 ⇒ 2 x 2 - ( 2 x ) 2 = 27 ⇒    2 x 2 = 27 ⇒    x = ± 3 3 2 ⇒ x 1 ,   x 2 = ± 3 3 2 Let M be the mid-point of R '   &   S ' So, the x coordinate of T is x 1 + x 2 2 = 0 It lies on y = 2 x + 1 So, the coordinates of T are 0 , 1 S