Question

Find the area of the smaller region bounded by the ellipse $\frac{x^2}{25}+\frac{y^2}{4}=1$ and the line $\frac{x}{5}+\frac{y}{2}=1$.

Answer 1

$\frac{x^2}{5^2}+\frac{y^2}{2^2}=1$ and $\frac{x}{5}+\frac{y}{2}=1$
$\Rightarrow y=\frac{2}{5}\sqrt{25-x^2}$
The points of intersections are $(0,2)$ and $(5,0)$.

Area of the required section = Area of ellipse in $I$ quadrant - Area of $△ABO$
$=\underset{0}{\overset{5}{\int }}\frac{2}{5}\sqrt{25-x^2}-\frac{1}{2}\times 5\times 2$
$=\frac{2}{5}{[\frac{x}{2}\sqrt{25-x^2}+\frac{25}{2}{\sin }^{-1}\frac{x}{5}]}_0^5-5$
$=\frac{2}{5}(0+\frac{25}{2}\times \frac{\pi }{2}-0)-5$
$=\frac{5\pi }{2}-5\text{sq. units}$
$=5(\frac{\pi }{2}-1)\text{sq. units}$