Find the area of the ellipse : \(\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}=1\). Hence write area of \(\dfrac{x^2}{25}+\dfrac{y^2}{16}=1\)

Answer: Area \(=\pi ab\) sq units; for the given ellipse, \(20\pi\) sq units.

The ellipse is symmetric about both axes, so its area is 4 times the area in the first quadrant.

From the equation, \(y=\dfrac ba\sqrt{a^2-x^2}\) in the first quadrant, for \(0\le x\le a\).

Area \(=4\displaystyle\int_0^a\dfrac ba\sqrt{a^2-x^2}\,dx=\dfrac{4b}{a}\left[\dfrac x2\sqrt{a^2-x^2}+\dfrac{a^2}{2}\sin^{-1}\dfrac xa\right]_0^a\)

\(=\dfrac{4b}{a}\left[0+\dfrac{a^2}{2}\sin^{-1}1-0\right]=\dfrac{4b}{a}\cdot\dfrac{a^2}{2}\cdot\dfrac{\pi}{2}=\pi ab\) sq units.

Hence: for \(\dfrac{x^2}{25}+\dfrac{y^2}{16}=1\), \(a=5\), \(b=4\), so area \(=\pi(5)(4)=20\pi\) sq units.