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Find the area of the smaller region bounded by the ellipse \(\Large { \frac{ x^2}{ a^2}}+\Large { \frac{ y^2}{ b^2}} =\normalsize 1\) and the line \(\Large { \frac{ x}{ a}}+\Large { \frac{y}{ b}} =\normalsize1\)

1 Answer

The area of the smaller region bounded by the ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1,$and the line $\frac{x}{a}+\frac{y}{b}=1$, is represented by the shaded region in the fig.
Hence the required area is the area enclosed between the straight line and the ellipse.
To find the limits,the semi major axis extends from 0 to a and the x intercept of the straight line is also a.Hence the limits are 0 to a.
The required area is $A=\int_a^b[f(x)-g(x)]dx$
Here a=0,b=a;$f(x)=\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ and $g(x)=\frac{x}{a}+\frac{y}{b}=1$
$A=\int_0^a\frac{b}{a}\sqrt{a^2-x^2}dx-\frac{b}{a}\int_0^a(a-x)dx$
Integrating we get,
$A=\frac{b}{a}\left \{\begin{bmatrix}\frac{x}{2}\sqrt{a^2-x^2}+\frac{a^2}{2}\sin^{-1}\frac{x}{a}\end{bmatrix}-\begin{bmatrix}ax-\frac{x^2}{2}\end{bmatrix}\right\}_0^a$
On applying the limits we get,
$A=\frac{b}{a}\begin{bmatrix}\frac{a^2}{2}\big(\frac{\pi}{2}\big)-\big(a^2-\frac{a^2}{2}\big)\end{bmatrix}$
$\;\;\;=\frac{b}{a}\begin{bmatrix}\frac{a^2\pi}{4}-\frac{a^2}{2}\end{bmatrix}$
$\;\;\;=\frac{ba^2}{2a}\begin{bmatrix}\frac{\pi}{2}-1\end{bmatrix}$
$\;\;\;=\frac{ab}{2}\begin{bmatrix}\frac{\pi}{2}-1\end{bmatrix}=\frac{ab}{4}(\pi-2)sq.units.$
Hence the required area is $\frac{ab}{4}(\pi-2)sq. units.$
answered Dec 21, 2013 by yamini.v
edited Sep 16, 2014 by sreemathi.v
 

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