Area of the region bounded by the ellipse 9x^ | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(A) The given equation of the ellipse is $9x^2 + 16y^2 = 1$. This can be rewritten in the standard form $\frac{x^2}{(1/3)^2} + \frac{y^2}{(1/4)^2} = 1

Vedclass · Accepted Answer

$\frac{\pi}{48}$

Vedclass · Answer

(A) The given equation of the ellipse is $9x^2 + 16y^2 = 1$. This can be rewritten in the standard form $\frac{x^2}{(1/3)^2} + \frac{y^2}{(1/4)^2} = 1$. Here,$a = \frac{1}{3}$ and $b = \frac{1}{4}$. The total area of an ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ is given by $\pi ab$. The area in the first quadrant is $\frac{1}{4}$ of the total area. Area $= \frac{1}{4} \pi ab = \frac{1}{4} \pi \left(\frac{1}{3}\right) \left(\frac{1}{4}\right) = \frac{\pi}{48}$ sq. units. Thus,the correct option is $A$.

Area of the region bounded by the ellipse $9x^2 + 16y^2 = 1$ in the first quadrant is . . . . . . sq. units.

Similar Questions

The area of the region $R = \{(x, y) : 0 \le y \le \frac{27}{x}, 1 \le x \le 9\}$ is equal to:

The area of the region bounded by $y=2x-x^{2}$ and the $x$-axis is

Find the area of the region bounded by the curve $y^{2}=4x$ and the line $x=3$. (in $\sqrt{3}$)

The area of the region $\{ (x, y) : x^2 + y^2 \le 1 \le x + y \}$ is

The smaller area enclosed by the circle $x^{2}+y^{2}=4$ and the line $x+y=2$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module