ellipse

area of the greatest rectangle that can be inscribed in ellipse x2/a2+y2/b2=1

3 Answers

1
Rohan Ghosh ·

2ab

1
Rohan Ghosh ·

as area = 4absinθcosθ=2absin2θ
its max value=2ab

62
Lokesh Verma ·

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)

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