The ellipse \(\displaystyle\frac{x^{2}}{16}+\frac{y^{2}}{b^{2}}=1\), where b > 0, passes through the point \(\left(2, \displaystyle\frac{3\sqrt{3}}{2}\right)\). The area (in sq. units) of the region enclosed by the ellipse is
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → \(12\pi\)
Since the point lies on the ellipse, its coordinates satisfy the equation:
\( \displaystyle \frac{4}{16}+\frac{27}{4b^{2}} = 1 \)
\( \displaystyle \frac{27}{4b^{2}} = \frac{3}{4} \)
\(b^{2} = 9,\ \ b = 3\)
So the semi-axes are a = 4 and b = 3, and the area enclosed by an ellipse is \(\pi ab\):
\(A = \pi\times4\times3 = 12\pi\)