2ab
area of the greatest rectangle that can be inscribed in ellipse x2/a2+y2/b2=1
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3 Answers
Lokesh Verma
·2008-12-08 12:46:05
y=+b sin t and lower value is -b sin t
x=a sin t
dx=a cost dt
area = integral of y.dx limit from -a to a
∫2b(sin t)dx
=∫2b sint . a cos t dt
=∫ab sin2t
The limits are 0,pi
finally, you will get 2ab! (sorry if i made some stupid mistake.. but i was giving the essence of the solution)