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Q: The area of the region bounded by the ellipse \frac{x^{2}}{25}+\frac{y^{2}}{16}=1  is
A. 20π sq units
B. 20π2 sq units
C. 16π2 sq units
D. 25π sq units

Answers (1)

A)

The ellipse \frac{x^{2}}{25}+\frac{y^{2}}{16}=1

Ellipse has a symmetry with x axis and y axis

The required area can be calculated as

\\ =4 \int_{0}^{5} \frac{4}{5}\left(\sqrt{25-x^{2}}\right) d x \\ {\left[\int \sqrt{a^{2}-x^{2}} d x=\frac{x \sqrt{a^{2}-x^{2}}}{2}+\frac{a^{2}}{2} \sin ^{-1}\left(\frac{x}{a}\right)\right]} \\ =\frac{16}{5} \int_{0}^{5}\left(\sqrt{5^{2}-x^{2}}\right) d x

\\ =\frac{16}{5}\left[\frac{x \sqrt{5^{2}-x^{2}}}{2}+\frac{5^{2}}{2} \sin ^{-1}\left(\frac{x}{5}\right)\right]_{0}^{5} \\ =\frac{16}{5}\left(0-\frac{25 \pi}{4}-0-0\right) \\ =20 \pi \text { sq.units }

 

 

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infoexpert22

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