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Q.
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
Application of Integrals
Solution:
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,
$I=\int_{0}^{\alpha}\frac{b}{a}\sqrt{a^{2}-x^{2}} dx-\frac{\alpha^{2}}{2}$
$=\frac{b}{2a}\left[x\sqrt{a^{2}-x^{2}}+a^{2} sin^{-1} \frac{x}{a}\right]_{0}^{\alpha}-\frac{\alpha^{2}}{2}$
$=\frac{b}{2a}\left[\alpha\sqrt{a^{2}-\alpha^{2}}+a^{2} sin^{-1} \frac{\alpha}{a}\right]-\frac{\alpha^{2}}{2}$
$=\frac{b}{2a}\cdot\frac{ab}{\sqrt{a^{2}+b^{2}}}\cdot\frac{a^{2}}{\sqrt{a^{2}+b^{2}}}$
$+\frac{ab}{2} sin^{-1} \frac{b}{\sqrt{a^{2}+b^{2}}}-\frac{a^{2}b^{2}}{2\left(a^{2}+b^{2}\right)}$
$=\frac{ab}{2} tan^{-1}\frac{b}{a}$
Required area $= 8I$
$= 4ab tan^{-1} \frac{b}{a}$ sq. units
