The area (in sq. units) of the smaller region bounded by the ellipse \(\displaystyle\frac{x^{2}}{25}+\frac{y^{2}}{9}=1\) and the line \(\displaystyle\frac{x}{5}+\frac{y}{3}=1\) is
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → \(\displaystyle\frac{15\pi}{4}-\frac{15}{2}\)
The line passes through (5, 0) and (0, 3), the ends of the semi-axes, so the smaller region lies in the first quadrant, between the arc of the ellipse above and the line below. On the arc \(y = \frac{3}{5}\sqrt{25-x^{2}}\), and on the line \(y = 3-\frac{3x}{5}\):
\( \displaystyle A = \frac{3}{5}\int_{0}^{5}\sqrt{25-x^{2}}\,dx-\int_{0}^{5}\left(3-\frac{3x}{5}\right)dx \)
The first integral is
\( \displaystyle \frac{3}{5}\left[\frac{x}{2}\sqrt{25-x^{2}}+\frac{25}{2}\sin^{-1}\frac{x}{5}\right]_{0}^{5} = \frac{3}{5}\times\frac{25\pi}{4} = \frac{15\pi}{4} \)
The second is
\( \displaystyle \left[3x-\frac{3x^{2}}{10}\right]_{0}^{5} = 15-\frac{15}{2} = \frac{15}{2} \)
So
\( \displaystyle A = \frac{15\pi}{4}-\frac{15}{2} \)
Shortcut: The region is a quarter of the ellipse minus a right triangle: \(\frac{\pi ab}{4}-\frac{ab}{2} = \frac{15\pi}{4}-\frac{15}{2}\).