Question

The area common to the ellipses $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ and $\frac{x^2}{b^2}+\frac{y^2}{a^2}=1$, $0<b<a$ is

• A. ${\left(a+b\right)}^{2}ta{n}^{-1}\frac{b}{a}$
• B. ${\left(a+b\right)}^{2}ta{n}^{-1}\frac{a}{b}$
• C. $4abta{n}^{-1}\frac{b}{a}$
• D. $4abta{n}^{-1}\frac{a}{b}$

Answer 1

Answer: (C)

We have, $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ and $\frac{x^2}{b^2}+\frac{y^2}{a^2}=1$ both
are ellipse with centre $\left(0,0\right)$, vertex $\left(a,0\right)$, $\left(-a,0\right)$ and $\left(0,b\right)$, $\left(0,-b\right)$ respectively
The curves intersect at $\left(\pm \alpha ,\pm \alpha \right)$ where $\alpha =\frac{ab}{\sqrt{a^2+b^2}}$

The area of the shaded region,
$I={\int }_0^{\alpha }\frac{b}{a}\sqrt{a^2-x^2}dx-\frac{{\alpha }^2}{2}$
$=\frac{b}{2a}{\left[x\sqrt{a^2-x^2}+a^{2}si{n}^{-1}\frac{x}{a}\right]}_0^{\alpha }-\frac{{\alpha }^2}{2}$
$=\frac{b}{2a}\left[\alpha \sqrt{a^2-{\alpha }^2}+a^{2}si{n}^{-1}\frac{\alpha }{a}\right]-\frac{{\alpha }^2}{2}$
$=\frac{b}{2a}\cdot \frac{ab}{\sqrt{a^2+b^2}}\cdot \frac{a^2}{\sqrt{a^2+b^2}}$
$+\frac{ab}{2}si{n}^{-1}\frac{b}{\sqrt{a^2+b^2}}-\frac{a^2{b}^2}{2\left(a^2+b^2\right)}$
$=\frac{ab}{2}ta{n}^{-1}\frac{b}{a}$
Required area $=8I$
$=4abta{n}^{-1}\frac{b}{a}$ sq. units