Question

The area enclosed by the ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ is equal to
$$\begin{gathered} \left(\mathrm{a}\right){\mathrm{\pi }}^2\mathrm{ab} \\ \left(\mathrm{b}\right)\mathrm{\pi }\mathrm{ab} \\ \left(\mathrm{c}\right)\mathrm{\pi }{\mathrm{a}}^2\mathrm{b} \\ \left(\mathrm{d}\right)\mathrm{\pi }{\mathrm{ab}}^2 \end{gathered}$$

Answer 1

To find: area enclosed by the ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$



The whole area enclosed by the given ellipse = 4(Area of the region bounded by AOBA)

Thus,
$$\begin{gathered} \mathrm{Area}\mathrm{of}\mathrm{the}\mathrm{ellipse}=4{\int }_0^{a}ydx \\ =4{\int }_0^a\left(b\sqrt{1-\frac{x^2}{a^2}}\right)dx \\ =\frac{4b}{a}{\int }_0^a\left(\sqrt{a^2-x^2}\right)dx \\ =\frac{4b}{a}{\left(\frac{x}{2}\sqrt{a^2-x^2}+\frac{{\left(a\right)}^2}{2}{\sin }^{-1}\frac{x}{a}\right)}_0^a \\ =\frac{4b}{a}\left[\left(\frac{a}{2}\sqrt{a^2-a^2}+\frac{a^2}{2}{\sin }^{-1}\frac{a}{a}\right)-\left(\frac{0}{2}\sqrt{a^2-0^2}+\frac{a^2}{2}{\sin }^{-1}\frac{0}{a}\right)\right] \\ =\frac{4b}{a}\left[\left(0+\frac{a^2}{2}{\sin }^{-1}1\right)-\left(0+\frac{a^2}{2}{\sin }^{-1}0\right)\right] \\ =\frac{4b}{a}\left[\left(\frac{a^2}{2}\frac{\mathrm{\pi }}{2}\right)-\left(0\right)\right] \\ =\frac{4b}{a}\left(\frac{\mathrm{\pi }{a}^2}{4}\right) \\ =\mathrm{\pi }ab \\ \mathrm{Thus},\mathrm{Area}=\mathrm{\pi }ab\mathrm{sq}.\mathrm{units} \end{gathered}$$


Hence, the correct option is (b).