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प्रश्न
Find the area of elipse `x^2/a^2+y^2/b^2=1`
उत्तर
equation of elipse `x^2/a^2+y^2/b^2=1`
Clearly the area of elipse is 4 times the area of region OPQD as show in the figure.For the region.Limits of integration are x=0 and x=a
From the elipse
`y^2/b^2=1-x^2/a^2=(a^2-x^2)/a^2`
`therefore y^2=b^2/a^2(a^2-x^2)`
`y=+-b/a(a^2-x^2)^(1/2)`
`y=b/a(a^2-x^2)^(1/2)`
We know
`A=4int_0^1ydx`
`=int_0^ab/a(a^2-x^2)^(1/2)dx`
`=(4b)/a[x/2(a^2-x^2)^(1/2)+a^2/2sin^(-1)(x/a)]_0^a`
`=(4b)/a[a^2/2 pi/2-0]`
`=ab pi `
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