Area of the ellipse frac{x^2}{a^2} + frac{y^2 | 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 $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$.Since the equation contains only even powers of $x$ and $y$,the ellip

Vedclass · Accepted Answer

$\pi ab 	ext{ sq. unit}$

Vedclass · Answer

(A) The given equation of the ellipse is $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$.Since the equation contains only even powers of $x$ and $y$,the ellipse is symmetric about both the $x$-axis and the $y$-axis.Therefore,the total area of the ellipse is $4$ times the area of the region in the first quadrant (region $OBC$).Area $= 4 \int_{0}^{a} y \, dx = 4 \int_{0}^{a} \frac{b}{a} \sqrt{a^2 - x^2} \, dx$.Let $x = a \sin 	heta$,then $dx = a \cos 	heta \, d	heta$. When $x=0, 	heta=0$ and when $x=a, 	heta=\frac{\pi}{2}$.Area $= 4 \frac{b}{a} \int_{0}^{\pi/2} \sqrt{a^2 - a^2 \sin^2 	heta} \cdot a \cos 	heta \, d	heta = 4ab \int_{0}^{\pi/2} \cos^2 	heta \, d	heta$.Using the identity $\cos^2 	heta = \frac{1 + \cos 2	heta}{2}$,we get:Area $= 4ab \int_{0}^{\pi/2} \frac{1 + \cos 2	heta}{2} \, d	heta = 2ab \left[ 	heta + \frac{\sin 2	heta}{2} ight]_{0}^{\pi/2}$.Area $= 2ab \left( (\frac{\pi}{2} + 0) - (0 + 0) ight) = \pi ab 	ext{ sq. unit}$.

Area of the ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ is

Similar Questions

Find the area of the region bounded by $x^{2}=4y$,$y=2$,$y=4$ and the $y-$axis in the first quadrant.

The area bounded by the curve $y^2 = 4ax$,the $x$-axis,and the ordinates $x = 0$ and $x = a$ is

The area between the curve $y = 4 + 3x - x^2$ and the $x$-axis is:

The area of the smaller part between the circle $x^2 + y^2 = 4$ and the line $x = 1$ is

The area bounded by the parabola $y = 4x^2$,the $y$-axis,and the lines $y = 1$ and $y = 4$ is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module