If the points of intersection of two distinct conics x 2 + y 2 = 4 b and x 2 16 + y 2 b 2 = 1 lie on the curve y 2 = 3 x 2 , then 3 3 times the area of the rectangle formed by the intersection points is _______.

Step-by-step solution

Given conics are x 2 + y 2 = 4 b and x 2 16 + y 2 b 2 = 1 . The intersection point of these conics lie on y 2 = 3 x 2 . So, putting y 2 = 3 x 2 in both the conics. ⇒ x 2 + 3 x 2 = 4 b , x 2 16 + 3 x 2 b 2 = 1 ⇒ x 2 = b , b 16 + 3 b b 2 = 1 ⇒ b 16 + 3 b = 1 ⇒ b 2 + 48 = 16 b ⇒ b 2 - 16 b + 48 = 0 ⇒ b 2 - 12 b - 4 b + 48 = 0 ⇒ b - 4 b - 12 ⇒ b = 4 , 12 ( b = 4 is rejected because curves coincide) ⇒ b = 12 ⇒ x = ± 12 ⇒ 12 + y 2 = 48 ⇒ y = ± 6 Hence points of intersection are ( ± 12 , ± 6 ) . So, length and breadth of