Question

Area enclosed by the curve \(\pi\left[4(x-\sqrt{2})^2+y^2\right]=8\) is

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

Hint:

For ellipse \(\frac{(x-h)^2}{a^2}+\frac{(y-k)^2}{b^2}=1\), area \(=\pi ab\).

Answer 1

The Correct Option is D

Step 1: Simplify the given curve equation.
\[ \pi\left[4(x-\sqrt{2})^2+y^2\right]=8 \Rightarrow 4(x-\sqrt{2})^2+y^2=\frac{8}{\pi} \]
Step 2: Write in standard ellipse form.
Divide both sides by \(\frac{8}{\pi}\):
\[ \frac{4(x-\sqrt{2})^2}{\frac{8}{\pi}}+\frac{y^2}{\frac{8}{\pi}}=1 \]
\[ \frac{(x-\sqrt{2})^2}{\frac{2}{\pi}}+\frac{y^2}{\frac{8}{\pi}}=1 \]
Step 3: Identify semi-axes.
So ellipse has:
\[ a^2=\frac{2}{\pi}, \quad b^2=\frac{8}{\pi} \]
\[ a=\sqrt{\frac{2}{\pi}}, \quad b=\sqrt{\frac{8}{\pi}} \]
Step 4: Area of ellipse.
\[ \text{Area}=\pi ab \]
\[ =\pi\left(\sqrt{\frac{2}{\pi}}\right)\left(\sqrt{\frac{8}{\pi}}\right) =\pi\sqrt{\frac{16}{\pi^2}} =\pi\left(\frac{4}{\pi}\right)=4 \]
Final Answer:
\[ \boxed{4} \]