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

Given the curve $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ ...(i)

and equation of line $\frac{x}{a}+\frac{y}{b}=1$ ...(ii)

On putting the value of $\frac{x}{a}$ from Eq. (ii)

in Eq. (i), we get

$(1-\frac{y}{b}{)}^2+\frac{y^2}{b^2}=1\Rightarrow 1+2(\frac{y^2}{b^2})-\frac{2y}{b}=1\Rightarrow \frac{2y}{b}(\frac{y}{b}-1)=0\Rightarrow y=0,y=bthenx=a,x=0$
i.e., the intersection points are A(a, 0) and B(0, b).
The required area is shown in the shaded figure. For the ellipse,
$\frac{y^2}{b^2}=1-\frac{x^2}{a^2}\Rightarrow |y|=\frac{b}{a}\sqrt{a^2-x^2}$
Now, area of $\Delta AOB=\frac{1}{2}|OA|.|OB|=\frac{1}{2}$ ab sq unit
Also, area under the ellipse in the first quadrant
$={\int }_0^{a}ydx={\int }_0^a\frac{b}{a}\sqrt{a^2-x^2}dx=\frac{b}{a}[\frac{x\sqrt{a^2-x^2}}{2}+\frac{a^2}{2}si{n}^{-1}\frac{x}{a}{]}_0^a=\frac{b}{2a}[(0+a^{2}si{n}^{-1}(1\left)\right)-(0+a^{2}si{n}^{-1}(0\left)\right)]=\frac{b}{2a}[a^2.\frac{\pi }{2}-a^2.0]=\frac{\pi ab}{4}squnit$
$\therefore$ Required area = Area of shaded region
= Area of curve OABO $-Areaof\Delta OAB=\frac{\pi ab}{4}-\frac{1}{2}ab=\frac{(\pi -2)ab}{4}=\frac{ab}{4}(\pi -2)squnit$