If the area enclosed between the curves y^2 = 4kx and y = kx, k > 0 is 2k 3 sq. units then k =

Step-by-step solution

Intersection points of y^2 = 4kx and y = kx are obtained by substituting y = kx into y^2 = 4kx . (kx)^2 = 4kx k^2x^2 - 4kx = 0 kx(kx - 4) = 0 Since k > 0 , x = 0 or x = 4 k . The area enclosed between the curves is given by: A = ₀^ 4/k (2 kx - kx) dx A = [ 4 3 k x^ 3/2 - k 2 x^2 ]₀^ 4/k A = 4 3 k ( 4 k )^ 3/2 - k 2 ( 4 k )^2 A = 32 3k - 8 k = 8 3k Given that the area is 2k 3 sq. units: 8 3k = 2k 3 2k^2 = 8 k^2 = 4 Since k > 0 , k = 2 . Answer: 2