# Find the area of the smaller region bounded by the ellipse

Question:

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

Solution:

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 BCAB as

∴ Area BCAB = Area (OBCAO) – Area (OBAO)

$=\int_{0}^{a} b \sqrt{1-\frac{x^{2}}{a^{2}}} d x-\int_{0}^{a} b\left(1-\frac{x}{a}\right) d x$

$=\frac{b}{a} \int_{0}^{a} \sqrt{a^{2}-x^{2}} d x-\frac{b}{a} \int_{b}^{a}(a-x) d x$

$=\frac{b}{a}\left[\left\{\frac{x}{2} \sqrt{a^{2}-x^{2}}+\frac{a^{2}}{2} \sin ^{-1} \frac{x}{a}\right\}_{0}^{a}-\left\{a x-\frac{x^{2}}{2}\right\}_{0}^{a}\right]$

$=\frac{b}{a}\left[\left\{\frac{a^{2}}{2}\left(\frac{\pi}{2}\right)\right\}-\left\{a^{2}-\frac{a^{2}}{2}\right\}\right]$

$=\frac{b}{a}\left[\frac{a^{2} \pi}{4}-\frac{a^{2}}{2}\right]$

$=\frac{b a^{2}}{2 a}\left[\frac{\pi}{2}-1\right]$

$=\frac{a b}{2}\left[\frac{\pi}{2}-1\right]$

$=\frac{a b}{4}(\pi-2)$