The sum of squares of all possible values of k , for which area of the region bounded by the parabolas 2 y 2 = k x and k y 2 = 2 y − x is maximum, is equal to:

Step-by-step solution

Given: k y 2 = 2 y − x , 2 y 2 = k x Finding the point of intersection, ⇒ k y 2 = 2 y − 2 y 2 k ⇒ y = 0 and k y = 2 1 − 2 y k ⇒ k y + 4 y k = 2 ⇒ y = 2 k + 4 k ⇒ y = 2 k k 2 + 4 So, the required area is given by, A = ∫ 0 2 k k 2 + 4 y − k y 2 2 − 2 y 2 k ⋅ d y ⇒ A = y 2 2 − k 2 + 2 k ⋅ y 3 3 0 2 k k 2 + 4 ⇒ A = 2 k k 2 + 4 2 1 2 − k 2 + 4 2 k × 1 3 × 2 k k 2 + 4 ⇒ A = 1 6 × 4 × 1 k + 4 k 2 We know that, A M ≥ G M ⇒ k + 4 k 2 ≥ 2 ⇒ k + 4 k ≥ 4 So, area is maximum when k = 4 k . ⇒ k = 2 , − 2 Hence, the sum of square