We have y2=4a(x+a)...(i), a parabola with vertex (−a,0)
and y2=4b(x−a)...(ii), a parabola with vertex (a,0)
Solving (i) and (ii), we get y=±ab−a8b A=20∫b−aa8b((4by2+a)−(4by2−a))dy =20∫b−aa8b(2a−4ab(b−a)y2)dy =2[2ay−12abb−ay3]0b−aa8b =2[2a×b−aa8b−12abb−a(b−aa8b)3] =4a2b−a8b−34a2b−a8b=38a2b−a8b sq. units