If the area bounded by y 2 = 4 a x and x 2 = 4 a y is 64 3 square units, then the positive value of a is

Step-by-step solution

Point of intersection of the two given curves are 0,0 and 4 a , 4 a The area bounded = 2 ∫ 0 4 a x - x 2 4 a d x = 2 x 2 2 - x 3 12 a 0 4 a = 2 8 a 2 - 16 a 2 3 = 16 a 2 3 sq. units Given, 16 a 2 3 = 64 3 Hence, the positive value of a = 2