Question

$AOB$ is the 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 are chord $AB$ of the ellipse is

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

Answer 1

The correct option is D $\frac{ab(\pi -2)}{4}$

$\frac{x^2}{a^2}+\frac{y^2}{b^2}=1⟹\left(1\right)$

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

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

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

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

From (1) if $x=0$then$y=\pm b$

From (1) if $y=0$then$x=\pm a$

The required area

${\int }_0^{a}ydx-AreaofOAB=\frac{b}{a}{\int }_0^a\sqrt{{a^2-x}^2}dx-\frac{1}{2}ab$

$=\frac{b}{a}[\frac{x}{2}\sqrt{{a^2-x}^2}+\frac{a^2}{2}{sin}^{-1}\frac{x}{a}]-\frac{ab}{2}$

$=\frac{b}{a}[\frac{a^2}{2}\ast \frac{x}{2}]-\frac{ab}{2}$

$=\frac{\pi ab}{4}-\frac{ab}{2}$

$=\frac{(\pi -2)ab}{4}$