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  • Find the area of the smaller region bounded by the ellipse x^2/a^2+y^2/b^2=1 and the line x/a+y/b=1.

Q : 9  Find the area of the smaller region bounded by the ellipse  \small \frac{x^2}{a^2}+\frac{y^2}{b^2}=1  and the line  \small \frac{x}{a}+\frac{y}{b}=1.
 

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The area of the shaded region ACB is to be found 

The given ellipse and the line intersect at following points

\left ( 0,b \right )and\left ( a,0 \right )

\\\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1\\ \Rightarrow y=\frac{b}{a}\sqrt{a^{2}-x^{2}}

Y will always be positive since the shaded region lies above x axis

\\\frac{x}{a}+\frac{y}{b}=1\\ \Rightarrow y=\frac{b}{a}(a-x)

The required area is

\\\frac{b}{a}\int_{0}^{a}(\sqrt{a^{2}-x^{2}}-(a-x))dx\\ =\frac{b}{a}[\frac{x}{2}\sqrt{a^{2}-x^{2}}+\frac{a^{2}}{2}sin^{-1}\frac{x}{a}-ax+\frac{x^{2}}{2}]_{0}^{a}\\ =\frac{b}{a}[(\frac{a^{2}}{2}\times \frac{\pi }{2}-a^{2}+\frac{a^{2}}{2})]\\ =\frac{b}{a}(\frac{\pi a^{2}}{4}-\frac{a^{2}}{2})\\ =\frac{ab}{4}(\pi -2)units

 

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