The area (in sq. units) of the region enclosed by the ellipse \(2x^{2}+3y^{2}=6\) is
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → \(\sqrt{6}\,\pi\)
Dividing the equation by 6 gives the standard form:
\( \displaystyle \frac{x^{2}}{3}+\frac{y^{2}}{2} = 1 \)
So \(a^{2} = 3\) and \(b^{2} = 2\), and the semi-axes are
\(a = \sqrt{3},\ \ b = \sqrt{2}\)
The area enclosed by an ellipse is \(\pi ab\):
\(A = \pi\times\sqrt{3}\times\sqrt{2} = \sqrt{6}\,\pi\)
Why the other options are incorrect
Option (1) \(6\pi\) — Incorrect. This takes 3 and 2 as the semi-axes; they are \(a^{2}\) and \(b^{2}\), so the semi-axes are their square roots.
Option (3) \(4\sqrt{6}\,\pi\) — Incorrect. This uses the full lengths of the axes, \(2\sqrt{3}\) and \(2\sqrt{2}\), in place of the semi-axes.
Option (4) \(\displaystyle\frac{\sqrt{6}\,\pi}{4}\) — Incorrect. This is only the part of the ellipse in the first quadrant.