If the area enclosed between the curves y = kx² and x = ky² (k > 0) is 1 square unit. Then k is?

The correct option is A (1/√3)
Area bounded by y² = 4ax and x² = 4by, a, b ≠ 0 is 16ab/3
by using formula:
4a = 1/k, 4b = k, k > 0
Area = 16 * (1/(4k)) * (k/4) / 3
= 1
=> k² = 1/3
=> k = 1/√3.