The area of the smaller region bounded by the ellipse \(\displaystyle\frac{x^2}{16}+\displaystyle\frac{y^2}{9}= 1\) and the straight line \(3x + 4y = 12\) is:
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → \((3\pi-6)\) sq units
The line \(3x + 4y = 12\) meets the axes at (4, 0) and (0, 3), the ends of the semi-axes of the ellipse, so both points lie on the ellipse. The smaller region lies in the first quadrant, between the arc of the ellipse (above) and the line (below):
\( \displaystyle \text{Ellipse: } y = \frac{3}{4}\sqrt{16 - x^2},\qquad \text{Line: } y = \frac{3}{4}(4 - x) \)
\( \displaystyle \text{Area} = \frac{3}{4}\int_{0}^{4}\sqrt{16 - x^2}\,dx - \frac{3}{4}\int_{0}^{4}(4 - x)\,dx \)
Using \(\int\sqrt{a^2 - x^2}\,dx = \frac{x}{2}\sqrt{a^2 - x^2} + \frac{a^2}{2}\sin^{-1}\frac{x}{a} + C\) with a = 4:
\( \displaystyle \int_{0}^{4}\sqrt{16 - x^2}\,dx = \left[\frac{x}{2}\sqrt{16 - x^2} + 8\sin^{-1}\frac{x}{4}\right]_{0}^{4} \)
\( \displaystyle = 8 \times \frac{\pi}{2} = 4\pi \)
\( \displaystyle \int_{0}^{4}(4 - x)\,dx = \left[4x - \frac{x^2}{2}\right]_{0}^{4} = 16 - 8 = 8 \)
\( \displaystyle \text{Area} = \frac{3}{4}(4\pi) - \frac{3}{4}(8) = 3\pi - 6 \)
Shortcut: the region is a quarter of the ellipse minus a right triangle.
\( \displaystyle \text{Quarter ellipse} = \frac{\pi ab}{4} \)
\( \displaystyle = \frac{\pi \times 4 \times 3}{4} = 3\pi \)
\( \displaystyle \text{Triangle} = \frac{1}{2} \times 4 \times 3 = 6 \)
So the area is \(3\pi - 6\).