If the area (in sq. units) of the region enclosed by the ellipse \(\displaystyle\frac{x^{2}}{a^{2}}+\frac{y^{2}}{9}=1\), where a > 0, is \(12\pi\), then a is equal to
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → 4
The ellipse has semi-axes a and 3. It is symmetric about both axes, so its area is four times the area in the first quadrant, where \(y = \frac{3}{a}\sqrt{a^{2}-x^{2}}\):
\( \displaystyle A = 4\times\frac{3}{a}\int_{0}^{a}\sqrt{a^{2}-x^{2}}\,dx \)
\( \displaystyle = \frac{12}{a}\left[\frac{x}{2}\sqrt{a^{2}-x^{2}}+\frac{a^{2}}{2}\sin^{-1}\frac{x}{a}\right]_{0}^{a} \)
\( \displaystyle = \frac{12}{a}\times\frac{a^{2}}{2}\times\frac{\pi}{2} = 3\pi a \)
Setting the area equal to \(12\pi\):
\(3\pi a = 12\pi\)
\(a = 4\)
Shortcut: The area enclosed by an ellipse is \(\pi ab\), so \(\pi\times a\times3 = 12\pi\) gives a = 4.