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Question

Find the area of greater rectangle that can be inscribed in an ellipse x2a2+y2b2=1.

A
8absin2θ
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B
6absin2θ
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C
8absin2θ
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D
2absin2θ
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Solution

The correct option is A 8absin2θ
Given that
x2a2+y2b2=1

y2b2=1x2a2

y2=b2(a2x2a2)

y2=b2a2(a2x2)

y=b2a2(a2x2)

y=baa2x2

Required axis=a04baa2x2dx

=4baa0a2x2dx
=8absin2θ
Hence, the option (A) is correct answer.

1132334_708058_ans_e25f760d5a3147b4a16048d4a4d92b35.PNG

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