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

Answer 1

The area of the smaller region bounded by the ellipse,$\frac{x^2}{a^2}$+$\frac{y^2}{b^2}$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} 1-\frac{x^2}{a^2} \,dx$$$$-\int_{0}^{a} b(1-\frac{x}{a}) \,dx

=$\frac{b}{a}$

$\int_{0}^{a} \sqrt{a^2-x^2} \,dx$

-\frac{b}{a}$$\int_{0}^{a} (a-x) \,dx

=$\frac{b}{a}$[{$\frac{x}{2}\sqrt{a^2-x^2}$+$\frac{a^2}{2}\sin^{-1}\frac{x}{a}$}a0-{ax-$\frac{x^2}{2}$}a0]

=$\frac{b}{a}$[{$\frac{a^2}{2}$($\frac{π}{2}$)}-{a2-$\frac{a^2}{2}$}]

=$\frac{b}{a}$[$\frac{a^2π}{4}$-$\frac{a^2}{2}$]

=$\frac{ba^2}{2a}$[$\frac{π}{2}$-1]

=$\frac{ab}{2}$[$\frac{π}{2}$-1]

=$\frac{ab}{4}$(π-2)