AOB is the positive quadrant of the ellipse f | vedclass

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

Question

(D) The equation of the ellipse is $\frac{x^2}{5^2} + \frac{y^2}{3^2} = 1$. Here,$a=5$ and $b=3$. The area of the positive quadrant $OAB$ is $\frac{1}

Vedclass · Accepted Answer

$\frac{15}{4}(\pi-2)$

Vedclass · Answer

(D) The equation of the ellipse is $\frac{x^2}{5^2} + \frac{y^2}{3^2} = 1$. Here,$a=5$ and $b=3$. The area of the positive quadrant $OAB$ is $\frac{1}{4} 	imes \pi ab = \frac{1}{4} 	imes \pi 	imes 5 	imes 3 = \frac{15\pi}{4}$. The area of the triangle $OAB$ with vertices $(0,0), (5,0), (0,3)$ is $\frac{1}{2} 	imes 	ext{base} 	imes 	ext{height} = \frac{1}{2} 	imes 5 	imes 3 = \frac{15}{2}$. The area between the arc $AB$ and the chord $AB$ is the area of the quadrant minus the area of the triangle. Area $= \frac{15\pi}{4} - \frac{15}{2} = \frac{15}{4}(\pi - 2)$ sq. units.

$AOB$ is the positive quadrant of the ellipse $\frac{x^2}{25}+\frac{y^2}{9}=1$ in which $OA=5, OB=3$. The area between the arc $AB$ and the chord $AB$ of the ellipse in sq. units is

Similar Questions

The area of the bounded region enclosed by the curve $y=3-\left|x-\frac{1}{2}\right|-|x+1|$ and the $x-$axis is

The area of the region bounded by the curve $y = 2 - x - 3x^2$, the $X$-axis, the $Y$-axis, and the line $x = -2$ is

The area (in sq. units) bounded by the curves $y=(x+1)^2, y=(x-1)^2$ and the line $y=\frac{1}{4}$ is

Using integration,find the area of the region bounded by the triangle whose vertices are $(-1, 0)$,$(1, 3)$,and $(3, 2)$.

The area bounded by the parabola $y^{2}=16x$ and its latus rectum in the first quadrant is

Vedclass Test Series

Exam Paper Generator

Online Exam Module