A point P moves so that the sum of squares of its distances from the points 1 , 2 and - 2 , 1 is 14 . Let f x , y = 0 be the locus of P , which intersects the x -axis at the points A , B and the y -axis at the point C , D . Then the area of the quadrilateral A C B D is equal to

Step-by-step solution

Given, A point P moves so that the sum of squares of its distances from the points 1 , 2 and - 2 , 1 is 14 . So, x - 1 2 + y - 2 2 + x + 2 2 + y - 1 2 = 14 ⇒ x 2 + y 2 + x - 3 y - 2 = 0 Put x = 0 ⇒ y 2 - 3 y - 2 = 0 ⇒ y = 3 ± 17 2 Put y = 0 ⇒ x 2 + x - 2 = 0 x + 2 x - 1 = 0 ∴     A - 2 , 0 , B 1 , 0 , C 0 , 3 + 17 2 , D 0 , 3 - 17 2 Area formed by A C B D will be = 1 2 - 2 0 0 3 + 17 2 1 0 0 3 - 17 2 - 2 0 = 1 2 - 3 - 17 - 3 2 - 17 2 + 3 2 - 17 2 + 3 - 17 = 1 2 × 3