Sol. The two curves are:

y 2 = 4a (x + a) ... (i)

and, y 2 = –4a (x –b)... (ii)

Clearly, these two curves represent two parabolas having their vertices at (–a, 0) and (b, 0) respectively as shown in fig.

To find the coordinates of their points of intersection, we solve (i) and (ii) together. Solving these equations, we find that the two curves intersect at A (b –a, 2 ) and B (b –a, –2 ). We have to find the area of the shaded region in fig. Let us slice the region into horizontal strips. For the approximating rectangle shown in fig., we have

Length = PQ = (x 1 –x 2 ), Width = Δ y and Area = (x 2 –x 1 ) Δ y

The approximating rectangle can move vertically between B and A.

∴ required area =

=

= dy

= 2 dy

= 2

= 2

= 2

= (a + b) sq. units