Question

Area (in sq. units) of the region outside $\frac{x}{2}+\frac{y}{3}=1$ and inside the ellipse $\frac{x^2}{4}+\frac{y^2}{9}=1$ is

• A. $3(\pi -2)$
• B. $6(\pi -2)$
• C. $6(4-\pi )$
• D. $3(4-\pi )$

Answer 1

The correct option is B

$6(\pi -2)$

Explanation for the correct option:

Step 1. Draw the curves according to the question:

Given that $\frac{x^2}{4}+\frac{y^2}{9}=1$, $\frac{x^2}{4}+\frac{y^2}{9}=1$

$\Rightarrow a=2,b=3$

Now, Area of ellipse $=\mathrm{\pi ab}$

$=6\mathrm{\pi }$

Step 2. Required area $=$Area of ellipse $-$ Area of quadrilateral

$$\begin{gathered} =\mathrm{\pi }\times 2\times 3-\frac{1}{2}\times 6\times 4 \\ =6\mathrm{\pi }-12 \\ =6\left(\mathrm{\pi }-2\right) \end{gathered}$$

$\therefore$Area bounded by the given curves is $\mathbf{6}\left(\pi -2\right)$

Hence, Option ‘B’ is Correct.