Hello,

We have, y 2 = 4ax................ (1)

Also, x 2 = 4ay.........................(2)

So, x = y2/4a

Hence, (y 2 /4a) 2 = 4ay

So, y 4 = 64a 3 y

So, y 4 – 64a 3 y = 0

y[y 3 – (4a) 3 ] = 0

Hence,  y = 0, 4a
When y = 0, x = 0 and when y = 4a, x = 4a.
The points of intersection of (1) and (2) are (0, 0) and (4a, 4a).
The area of the region between the two curves

= 0 ∫ 4a (y1 – y 2 )dx

= 0 ∫ 4a [√(4ax) – x2/4a]dx

= [2√a.(x3/2)/(3/2) – (1/4a)(x 3 /3)]0 4a

= 4/3√a(4a)3/2 – (1/12a)(4a) 3 – 0

= 32/3a 2 – 16/3a 2

= 16/3a 2 sq. units.

Best Wishes.