# 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.

23 Points
14 years ago

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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