Question

The ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ is divided into two parts by the line $2x=a$. The area of the smaller part is

• A. $\left(\frac{\pi }{3}-\frac{\sqrt{3}}{2}\right)ab$ sq. units
• B. $\left(\frac{\pi }{3}+\frac{\sqrt{3}}{2}\right)ab$ sq. units
• C. $\left(\frac{\pi }{3}-\frac{\sqrt{3}}{4}\right)ab$ sq. units
• D. $\left(\frac{\pi }{3}+\frac{\sqrt{3}}{4}\right)ab$ sq. units

Answer 1

Answer: (C)

We have ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$, with centre $\left(0,0\right)$

Required area = area of shaded region
$A=2\cdot {\int }_{\frac{a}{2}}^{a}ydx$
$=\frac{2b}{a}{\int }_{\frac{a}{2}}^a\sqrt{a^2-x^2}dx$,
Put $x=asin\theta$
$\Rightarrow dx=acos\theta d\theta$
$\Rightarrow A=2ab{\int }_{\frac{\pi }{6}}^{\frac{\pi }{2}}co{s}^2\theta d\theta$
$=ab{\int }_{\frac{\pi }{6}}^{\frac{\pi }{2}}\left(1+cos2\theta \right)d\theta$
$=ab{\left[\theta +\frac{sin2\theta }{2}\right]}_{\frac{\pi }{6}}^{\frac{\pi }{2}}$
$=ab\left[\frac{\pi }{3}-\frac{\sqrt{3}}{4}\right]$ sq. units