find the greatest area of the rectangle which can be inscribed in the ellipse ; x2/a2+y2/b2=1 with sides parallel to the co-ordinate axes.

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

Please feel free to post as many doubts on our discussion forum as you can. If you find any question
Difficult to understand - post it here and we will get you the answer and detailed solution very quickly. We
are all IITians and here to help you in your IIT JEE preparation.

All the best  !!!

Regards,

Askiitians Experts

Adapa Bharath