The area of the smaller region bounded by ellipse \( \displaystyle\frac{x^2}{16}+\frac{y^2}{9}=1 \) and the straight line \( 3x+4y=12 \) is
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → \( (3\pi-6) \) Sq. units
The line \( 3x+4y=12 \) passes through (4, 0) and (0, 3), which are the ends of the semi-axes of the ellipse. The smaller region lies in the first quadrant, between the arc of the ellipse and this chord.
\( \displaystyle \text{Area}=\int_{0}^{4}\frac{3}{4}\sqrt{16-x^2}\,dx-\int_{0}^{4}\frac{3}{4}(4-x)\,dx \)
\( \displaystyle \int_{0}^{4}\sqrt{16-x^2}\,dx=\left[\frac{x}{2}\sqrt{16-x^2}+\frac{16}{2}\sin^{-1}\frac{x}{4}\right]_{0}^{4} \)
\( \displaystyle =8\cdot\frac{\pi}{2}=4\pi \)
\( \displaystyle \int_{0}^{4}(4-x)\,dx=\left[4x-\frac{x^2}{2}\right]_{0}^{4} \)
\( \displaystyle =16-8=8 \)
\( \displaystyle \text{Area}=\frac{3}{4}(4\pi)-\frac{3}{4}(8) \)
\( \displaystyle =3\pi-6 \)
Shortcut: A quarter of the ellipse has area \( \frac{1}{4}\pi ab=\frac{1}{4}\pi(4)(3)=3\pi \); the triangle with vertices (0, 0), (4, 0), (0, 3) has area 6. The smaller region is \( 3\pi-6 \).