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find the greastest area of the rectangle which can be inscribed in the ellipse; x2/a2+y2/b2=1 with sides parallel to the co-ordinate axes
let the vertices of rectangle be (asinø,bcosø),(asinø,-bcosø),(-asinø,-bcosø),(-asinø,bcosø).(in clockwise sense)
then area of rectangle =2absin2ø
d(area)/dø=2abcosø;
for(maxima) 2ab cosø=0
so ø=pi/4
also d2a/dø2 <0for(ø=pi/4)
hence area =2ab
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