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Question

The area enclosed by a curve with equation x2a2+y2b2=1 is

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Solution

x2a2+y2b2=1

y2b2=1x2a2=a2x2a2

y2=b2a2(a2x2)

y=baa2x2

Area =y dx
=2aabaa2x2dx

=4baa0a2x2dx

=4ba[x2a2x2+a22sin1(xa)]a0

=4ba[x2a2a2+a22sin1(a0)0a22sin10]

=4ba22aπ2=πab


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