Question

$AOB$ is a positive quadrant of the ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$, where $OA=a$, $OB=b$. The area between the arc $AB$ and chord $AB$ of the ellipse is

• A. $\pi ab$ sq. units
• B. $\left(\pi -2\right)ab$ sq. units
• C. $\frac{ab\left(\pi +2\right)}{2}$ sq. units
• D. $\frac{ab\left(\pi -2\right)}{4}$ sq. units

Answer 1

Answer: (D)

Required area
$=\underset{0}{\overset{a}{\int }}\left(\frac{b}{a}\sqrt{a^2-x^2}-\frac{b}{a}\left(a-x\right)\right)dx$

$=\frac{b}{a}\underset{0}{\overset{a}{\int }}\sqrt{a^2-x^2}dx-\frac{b}{a}\underset{0}{\overset{a}{\int }}\left(a-x\right)dx$
$=\frac{b}{a}{\left[\frac{x}{2}\sqrt{a^2-x^2}+\frac{a^2}{2}si{n}^{-1}\frac{x}{a}\right]}_0^a-\frac{b}{a}{\left[ax-\frac{x^2}{2}\right]}_0^a$

$=\frac{b}{a}\left[\frac{a^2}{2}si{n}^{-1}1\right]-\frac{b}{a}\left[a^2-\frac{a^2}{2}\right]$

$=\frac{ab}{2}\frac{\pi }{2}-ba\left(\frac{1}{2}\right)=\frac{ab}{4}\left(\pi -2\right)$ sq. units