The area (in sq. units) of the region in the first quadrant bounded by the ellipse \(\displaystyle\frac{x^{2}}{9}+\frac{y^{2}}{4}=1\), the x-axis and the line 2x = 3y is
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → \(\displaystyle\frac{3\pi}{4}\)
The line is \(\displaystyle y = \frac{2x}{3}\). It meets the ellipse where
\( \displaystyle \frac{x^{2}}{9}+\frac{1}{4}\times\frac{4x^{2}}{9} = 1 \)
\( \displaystyle \frac{2x^{2}}{9} = 1,\ \ x = \frac{3}{\sqrt{2}} \) (first quadrant)
The region lies above the x-axis. Its top is the line up to \(x = \displaystyle\frac{3}{\sqrt{2}}\) and the ellipse \(\displaystyle y = \frac{2}{3}\sqrt{9-x^{2}}\) from there to x = 3, so the area is found in two parts.
Under the line:
\( \displaystyle A_{1} = \int_{0}^{3/\sqrt{2}}\frac{2x}{3}\,dx = \left[\frac{x^{2}}{3}\right]_{0}^{3/\sqrt{2}} = \frac{3}{2} \)
Under the ellipse:
\( \displaystyle A_{2} = \frac{2}{3}\left[\frac{x}{2}\sqrt{9-x^{2}}+\frac{9}{2}\sin^{-1}\frac{x}{3}\right]_{3/\sqrt{2}}^{3} \)
\( \displaystyle = \frac{2}{3}\left[\frac{9\pi}{4}-\left(\frac{9}{4}+\frac{9\pi}{8}\right)\right] \)
\( \displaystyle = \frac{2}{3}\left(\frac{9\pi}{8}-\frac{9}{4}\right) = \frac{3\pi}{4}-\frac{3}{2} \)
Adding the two parts:
\( \displaystyle A = \frac{3}{2}+\frac{3\pi}{4}-\frac{3}{2} = \frac{3\pi}{4} \)
Why the other options are incorrect
Option (1) \(\displaystyle\frac{3\pi}{2}\) — Incorrect. This is the whole first-quadrant quarter of the ellipse, \(\displaystyle\frac{1}{4}\pi\times3\times2\); the line cuts off only part of it.
Option (2) \(\displaystyle\frac{3\pi}{8}\) — Incorrect. The region is half of the first-quadrant quarter, \(\displaystyle\frac{1}{2}\times\frac{3\pi}{2} = \frac{3\pi}{4}\), not a quarter of it.
Option (4) \(\displaystyle\frac{3\pi}{4}-\frac{3}{2}\) — Incorrect. This is only the part under the ellipse; the triangle of area \(\displaystyle\frac{3}{2}\) under the line is left out.