Question

The area enclosed by the ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ is equal to

• A. ${\pi }^2\mathrm{ab}$
• B. $\mathrm{\pi ab}$
• C. ${\mathrm{\pi a}}^{2}b$
• D. ${\mathrm{\pi ab}}^2$

Answer 1

The correct option is B $\mathrm{\pi ab}$

Area of the ellipse
=4 x Area of bounded by ellipse in each quadrant
$=4\times {\int }_0^{a}y\mathrm{dx}=4\times {\int }_0^a\frac{b}{a}\sqrt{a^2-x^2}\mathrm{dx}=\frac{4b}{a}{\left(\frac{x}{2}\sqrt{a^2-x^2}+\frac{a^2}{2}{\sin }^{-1}\frac{x}{a}\right]}_0^a=\frac{4b}{a}\left(\frac{{\mathrm{\pi a}}^2}{4}\right)=\mathrm{\pi ab}$