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

$=\frac{b}{a} \int\limits_{0}^{a}\sqrt{a^{2}-x^{2}} dx-\frac{b}{a} \int\limits_{0}^{a}\left(a-x\right)dx$
$=\frac{b}{a}\left[\frac{x}{2}\sqrt{a^{2}-x^{2}}+\frac{a^{2}}{2}sin^{-1}\frac{x}{a}\right]_{0}^{a}-\frac{b}{a}\left[ax-\frac{x^{2}}{2}\right]_{0}^{a}$

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

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