The area (in sq. units) of the region enclosed by the curve \(9x^2 +4y^2 = 36\) is
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → 6π
Dividing by 36 gives the standard form of an ellipse:
\( \displaystyle \frac{x^2}{4} + \frac{y^2}{9} = 1 \)
Its semi-axes are 2 (along the x-axis) and 3 (along the y-axis). The area of an ellipse with semi-axes a and b is \(\pi ab\):
\(\text{Area} = \pi \times 2 \times 3 = 6\pi\)
This follows from integration: the ellipse is symmetric about both axes, and in the first quadrant \(y = \frac{3}{2}\sqrt{4 - x^2}\), so
\( \displaystyle \text{Area} = 4\int_{0}^{2}\frac{3}{2}\sqrt{4 - x^2}\,dx = 6\int_{0}^{2}\sqrt{4 - x^2}\,dx \)
The last integral is the area of a quarter circle of radius 2, which is \(\pi\); so the area is \(6\pi\).
NCERT (2026-27 edition):
Area Under Simple Curves
