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 $\frac{a b}{4}(\pi-2)$ 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
$\frac{x^2}{9}+\frac{y^2}{4}=1 .....$(i)
and equation of line is
$\frac{x}{3}+\frac{y}{2}=1 ....$(ii)
For the points of intersection of ellipse and line, put the value of $x$ from Eq. (ii) in Eq. (i), we get
$\left(\frac{y}{2}-1\right)^2+\frac{y^2}{4}=1$

$ \Rightarrow \frac{y^2}{4}+1-y+\frac{y^2}{4}=1 \Rightarrow y^2-2 y=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
$=\left(\right.$ 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 $x=3$ )
$=\int\limits_0^3 2 \sqrt{1-\frac{x^2}{9}} d x-\int\limits_0^3 2\left(1-\frac{x}{3}\right) d x$
$=\frac{2}{3} \int\limits_0^3 \sqrt{3^2-x^2} d x-\frac{2}{3} \int\limits_0^3(3-x) d x$
$=\frac{2}{3}\left[\frac{x}{2} \sqrt{3^2-x^2}+\frac{9}{2} \sin ^{-1} \frac{x}{3}\right]_0^3-\frac{2}{3}\left[3 x-\frac{x^2}{2}\right]_0^3$
$\left(\because \int \sqrt{a^2-x^2} d x=\frac{x}{2} \sqrt{a^2-x^2}+\frac{a^2}{2} \sin ^{-1} \frac{x}{a}\right)$
$=\frac{2}{3}\left[0+\frac{9}{2} \sin ^{-1}(1)-0\right]-\frac{2}{3}\left[9-\frac{9}{2}-0\right]$
$=3\left(\frac{\pi}{2}\right)-(3)=3\left(\frac{\pi}{2}-1\right)$
$=\frac{3}{2}(\pi-2)$ sq units
Reason (R) Given, the curve is
$\frac{x^2}{a^2}+\frac{y^2}{b^2}=1 ....$(i)
and equation of line is
$\frac{x}{a}+\frac{y}{b}=1.....$(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{2 y}{b}=1 $
$ \Rightarrow \frac{2 y}{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 $\Delta A O B=\frac{1}{2} \cdot|O A| \cdot|O B|=\frac{1}{2} a b$ sq unit
Also, area under the ellipse in the first quadrant
$ =\int\limits_0^a y d x=\int\limits_0^a \frac{b}{a} \sqrt{a^2-x^2} d x $
$ =\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}{2 a}\left[\left(0+a^2 \sin ^{-1}(1)\right)-\left(0+a^2 \sin ^{-1}(0)\right)\right]$
$ =\frac{b}{2 a}\left[a^2 \cdot \frac{\pi}{2}-a^2 \cdot 0\right]=\frac{\pi a b}{4} \text { sq unit }$
$\therefore$ Required area $=$ Area of shaded region
$=$ Area of curve OABO - Area of $\triangle O A B$
$ =\frac{\pi a b}{4}-\frac{1}{2} a b=\frac{(\pi-2) a b}{4}$
$ =\frac{a b}{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