Dear Upendra,
Consider a rectangle as shown in the figure inscribed in the given ellipse
(x/a)2 + (y/b)2 = 1 such that its side are of length 2p and 2q
Now considerthe point (p,q) which is one of the ocrners of the rectangle that lies on the ellipse so
(p/a)2 + (q/b)2 = 1   which gives q= b * √(1-(p/a)2
now area of the rectangle A = 2p * 2q
                                        = 4 * p* q
                                        = 4 * p * b * √(1-(p/a)2
Now to find maximum area we find the maxima by differentiating it and equating it to zero
(dA/da) = 0
4*b*√(1-(p/a)2 + (4*p*b/√(1-(p/a)2) * (-p/a2) = 0
p = a/√2 and q = b/√2
Thus max area is found by substituting these vaues and comes out to be 2* a * b

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