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)