The area (in sq. units) of the smaller region cut off from the ellipse \(\displaystyle\frac{x^{2}}{16}+\frac{y^{2}}{4}=1\) by the line x = 2 is
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → \(\displaystyle\frac{8\pi}{3}-2\sqrt{3}\)
On the ellipse \(y = \pm\frac{1}{2}\sqrt{16-x^{2}}\). The smaller region lies between x = 2 and x = 4, and it is symmetric about the x-axis, so
\( \displaystyle A = 2\times\frac{1}{2}\int_{2}^{4}\sqrt{16-x^{2}}\,dx \)
\( \displaystyle = \left[\frac{x}{2}\sqrt{16-x^{2}}+8\sin^{-1}\frac{x}{4}\right]_{2}^{4} \)
At x = 4 the bracket is \(8\times\frac{\pi}{2} = 4\pi\). At x = 2 it is \(1\times\sqrt{12}+8\times\frac{\pi}{6} = 2\sqrt{3}+\frac{4\pi}{3}\). So
\( \displaystyle A = 4\pi-2\sqrt{3}-\frac{4\pi}{3} \)
\( \displaystyle = \frac{8\pi}{3}-2\sqrt{3} \)
Why the other options are incorrect
Option (1) \(\displaystyle\frac{4\pi}{3}-\sqrt{3}\) — Incorrect. This is only the half above the x-axis.
Option (2) \(\displaystyle\frac{16\pi}{3}-4\sqrt{3}\) — Incorrect. This is the segment cut off from the circle \(x^{2}+y^{2}=16\) by x = 2; the factor \(\frac{b}{a} = \frac{1}{2}\) of the ellipse is left out.
Option (3) \(\displaystyle\frac{16\pi}{3}+2\sqrt{3}\) — Incorrect. This is the larger region: the ellipse encloses \(\pi\times4\times2 = 8\pi\), and \(8\pi-\left(\frac{8\pi}{3}-2\sqrt{3}\right) = \frac{16\pi}{3}+2\sqrt{3}\).