Question

Find the area of the smaller region bounded by the ellipse $\frac{x^2}{9}+\frac{y^2}{4}=1$ and the line $\frac{x}{3}+\frac{y}{2}=1.$

Answer 1

Given the curve (represents an elliplse with centre at (0, 0)

$\frac{x^2}{9}+\frac{y^2}{4}=1$ ...(i)

and equation of line $\frac{x}{3}+\frac{y}{2}=1$ ...(ii)

For the points of intersection of ellipse and line put the value of x from

Eq. (ii) in Eq. (i). we get $(\frac{y}{2}-1{)}^2+\frac{y^2}{4}=1$

$\Rightarrow \frac{y^2}{4}+1-y+\frac{y^2}{4}=1\Rightarrow y^2-2y=0\Rightarrow y=0,2$

When y = 0, x = 3 and y = 2, x= 0

i.e., intersection points are A(3, 0) and B(0, 2)

$\therefore$ Required area = (Area under the curve $\frac{x^2}{9}+\frac{y^2}{4}=1,$ between x = 0 and x = 3) - (Area under the line $\frac{x}{3}+\frac{y}{2}=1,$ between x = 0 and x = 3)

$={\int }_0^{3}2\sqrt{1-\frac{x^2}{9}}dx-{\int }_0^{3}2(1-\frac{x}{3})dx=\frac{2}{3}{\int }_0^3\sqrt{3^2-x^2}dx-\frac{2}{3}{\int }_0^3(3-x)dx=\frac{2}{3}[\frac{x}{2}\sqrt{3^2-x^2}+\frac{9}{2}si{n}^{-1}\frac{x}{3}{]}_0^3-\frac{2}{3}[3x-\frac{x^2}{2}{]}_0^3$
$[\because \int \sqrt{a^2-x^2}dx=\frac{x}{2}\sqrt{a^2-x^2}+\frac{a^2}{2}si{n}^{-1}\frac{x}{a}]$

$=\frac{2}{3}[0+\frac{9}{2}si{n}^{-1}(1)-0]-\frac{2}{3}[9-\frac{9}{2}-0]=3\left(\frac{\pi }{2}\right)-\left(3\right)=3(\frac{\pi }{2}-1)=\frac{3}{2}(\pi -2)squnit.$

$=\frac{2}{3}[0+\frac{9}{2}si{n}^{-1}(1)-0]-\frac{2}{3}[9-\frac{9}{2}-0]=3\left(\frac{\pi }{2}\right)-\left(3\right)=3(\frac{\pi }{2}-1)=\frac{3}{2}(\pi -2)squnit.$