Question

The area enclosed by a curve with equation $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ is

Answer 1

$\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$

$\frac{y^2}{b^2}=1-\frac{x^2}{a^2}=\frac{a^2-x^2}{a^2}$

$y^2=\frac{b^2}{a^2}(a^2-x^2)$

$y=\frac{b}{a}\sqrt{a^2-x^2}$

Area $=\int ydx$

$=2{\int }_{-a}^a\frac{b}{a}\sqrt{a^2-x^2}dx$

$=\frac{4b}{a}{\int }_0^a\sqrt{a^2-x^2}dx$

$=\frac{4b}{a}{[\frac{x}{2}\sqrt{a^2-x^2}+\frac{a^2}{2}{\sin }^{-1}\left(\frac{x}{a}\right)]}_0^a$

$=\frac{4b}{a}[\frac{x}{2}\sqrt{a^2-a^2}+\frac{a^2}{2}{\sin }^{-1}\left(\frac{a}{0}\right)-0-\frac{a^2}{2}{\sin }^{-1}0]$

$=\frac{4b{a}^2}{2a}\frac{\pi }{2}=\pi ab$