Question

Assertion (A) The area of the smaller region bounded by the ellipse $\frac{x^2}{9}+\frac{y^2}{4}=1$ and the line $\frac{x}{3}+\frac{y}{2}=1$ is $\frac{3}{2}(\pi -2)$ sq units.
Reason (R) Formula to calculate the area of the smaller region bounded by the ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ and the line $\frac{x}{a}+\frac{y}{b}=1$ is $\frac{ab}{4}(\pi -2)$ sq units.

• A. Both A and R are correct; R is the correct explanation of A
• B. Both A and R are correct; $R$ is not the correct explanation of $A$
• C. $A$ is correct; $R$ is incorrect
• D. $R$ is correct; $A$ is incorrect

Answer 1

Answer: (A)

Assertion (A) Given curve, represents an ellipse with centre at $(0,0)$, is
$\frac{x^2}{9}+\frac{y^2}{4}=\mathrm{1.....}$(i)
and equation of line is
$\frac{x}{3}+\frac{y}{2}=\mathrm{1....}$(ii)
For the points of intersection of ellipse and line, put the value of $x$ from Eq. (ii) in Eq. (i), we get
${\left(\frac{y}{2}-1\right)}^2+\frac{y^2}{4}=1$

$\Rightarrow \frac{y^2}{4}+1-y+\frac{y^2}{4}=1\Rightarrow y^2-2y=0$
$\Rightarrow y=0,2$
When $y=0$, then $x=3$ and when $y=2$, then $x=0$
Thus, 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 $x=3$ )
$=\underset{0}{\overset{3}{\int }}2\sqrt{1-\frac{x^2}{9}}dx-\underset{0}{\overset{3}{\int }}2\left(1-\frac{x}{3}\right)dx$
$=\frac{2}{3}\underset{0}{\overset{3}{\int }}\sqrt{3^2-x^2}dx-\frac{2}{3}\underset{0}{\overset{3}{\int }}(3-x)dx$
$=\frac{2}{3}{\left[\frac{x}{2}\sqrt{3^2-x^2}+\frac{9}{2}{\sin }^{-1}\frac{x}{3}\right]}_0^3-\frac{2}{3}{\left[3x-\frac{x^2}{2}\right]}_0^3$
$\left(\because \int \sqrt{a^2-x^2}dx=\frac{x}{2}\sqrt{a^2-x^2}+\frac{a^2}{2}{\sin }^{-1}\frac{x}{a}\right)$
$=\frac{2}{3}\left[0+\frac{9}{2}{\sin }^{-1}(1)-0\right]-\frac{2}{3}\left[9-\frac{9}{2}-0\right]$
$=3\left(\frac{\pi }{2}\right)-(3)=3\left(\frac{\pi }{2}-1\right)$
$=\frac{3}{2}(\pi -2)$ sq units
Reason (R) Given, the curve is
$\frac{x^2}{a^2}+\frac{y^2}{b^2}=\mathrm{1....}$(i)
and equation of line is
$\frac{x}{a}+\frac{y}{b}=\mathrm{1.....}$(ii)

On putting the value of $\frac{x}{a}$ from Eq. (ii) in Eq. (i), we get
${\left(1-\frac{y}{b}\right)}^2+\frac{y^2}{b^2}=1$
$\Rightarrow 1+2\left(\frac{y^2}{b^2}\right)-\frac{2y}{b}=1$
$\Rightarrow \frac{2y}{b}\left(\frac{y}{b}-1\right)=0$
When $y=0$ and $y=b$, then $x=a$ and $x=0$, respectively.
Thus, 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 $\Delta AOB=\frac{1}{2}\cdot |OA|\cdot |OB|=\frac{1}{2}ab$ sq unit
Also, area under the ellipse in the first quadrant
$=\underset{0}{\overset{a}{\int }}ydx=\underset{0}{\overset{a}{\int }}\frac{b}{a}\sqrt{a^2-x^2}dx$
$=\frac{b}{a}{\left[\frac{x\sqrt{a^2-x^2}}{2}+\frac{a^2}{2}{\sin }^{-1}\frac{x}{a}\right]}_0^a$
$=\frac{b}{2a}\left[\left(0+a^2{\sin }^{-1}(1)\right)-\left(0+a^2{\sin }^{-1}(0)\right)\right]$
$=\frac{b}{2a}\left[a^2\cdot \frac{\pi }{2}-a^2\cdot 0\right]=\frac{\pi ab}{4}\text{ sq unit }$
$\therefore$ Required area $=$ Area of shaded region
$=$ Area of curve OABO - Area of $△OAB$
$=\frac{\pi ab}{4}-\frac{1}{2}ab=\frac{(\pi -2)ab}{4}$
$=\frac{ab}{4}(\pi -2)\text{ sq unit }$
If we put $a=3$ and $b=2$ in above area, then area enclosed by $\frac{x^2}{9}+\frac{y^2}{4}=1$ and $\frac{x}{3}+\frac{y}{2}=1$ is $\frac{6}{4}(\pi -2)$
i.e., $\frac{3}{2}(\pi -2)$ sq units
Thus, R is correct explanation of A