The area included between the parabolas y² = 4a(x + a) and y² = 4b (b - x) is:

$\frac{8}{3}\sqrt{ab}(a+b)$

y² = 4a(x + a) or y² = 4b (b - x)

Eliminating y², we get : x co-ordinate of their point of intersection as x = b - a, $y = ±2\sqrt{ab}$

The vertices of the parabolas are (-a, 0),(0, b) and are as shown in the figure.

$A=2[\int_{-a}^{b-a}y_1dx+\int_{b-a}^{b}y_2dx]=2[2\sqrt{a}\int_{-a}^{b-a}\sqrt{x+a}dx+\int_{b-a}^{b}2\sqrt{b}\sqrt{b-x}dx]$

$4\sqrt{2}.\frac{2}{3}[(x+a)^{3/2}]_{-a}^{b-a}+4\sqrt{b}.(-\frac{2}{3})[(b-x)^{3/2}]_{b-a}^{b}$

$=\frac{8}{3}\sqrt{ab}.a+\frac{8}{3}\sqrt{ab}.b=\frac{8}{3}\sqrt{ab}(a+b)$