Given parabolas are y2 = 2px …. (i) and x2 = 2py …. (ii)
Now, from equation (ii) we have
y = x2/2p
Putting the value of y in equation (i), we have
(x2/2p)2 = 2px
x4/4p2 = 2px
x4 = 8p3x
x4 – 8p3x = 0
x(x3 – 8p3) = 0
So, x = 0 or x3 – 8p3 = 0 ⇒ x = 2p
Now, the required area is
Therefore, the required area is 4/3 p2 sq. units.