Find the area of ellipse \(\frac{x^2}{25}+\frac{y^2}{4}=1\)
Q.2 (B) (iii) Find the area of ellipse \(\frac{x^2}{25}+\frac{y^2}{4}=1\)
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Answer: \(10\pi\) sq. units
From the equation, \(y=\frac{2}{5}\sqrt{25-x^2}\) in the first quadrant. The ellipse is symmetric about both axes, so
Area \(=4\int_0^5\frac{2}{5}\sqrt{25-x^2}\,dx=\frac{8}{5}\left[\frac{x}{2}\sqrt{25-x^2}+\frac{25}{2}\sin^{-1}\frac{x}{5}\right]_0^5\)
\(=\frac{8}{5}\left[0+\frac{25}{2}\cdot\frac{\pi}{2}-0\right]=\frac{8}{5}\cdot\frac{25\pi}{4}=10\pi\) sq. units.
Doubtly AI
· 28 Oct 2026
AI-verified · Oct 2026
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