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 , 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 ^ 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 distance between R^ and S^ . The value of D is
A. A
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 Slope of the line perpendicular to y = 2 x + 1 is - 1 2 So, the equation of perpendicular bisector is y - 1 = - 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 ⇒ 7 ( y - 1 ) 2 = 27 ⇒ y - 1 = ± 3 3 7 ⇒ y = 1 ± 3 3 7 ⇒ q 1 ,   q 2 = 1 ± 3 3 7 So, q 2 - q 1 2 = 6 3 7 2 Hence, D = 5 q 2 - q 1 2 = 5 × 6 3 7 2 = 77 . 14
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