Answer:
Area of ellipse=[tex]\pi ab[/tex]
Step-by-step explanation:
We are given that
[tex]x=acos\theta[/tex]
[tex]y=bsin\theta[/tex]
[tex]0\leq\theta\leq 2\pi[/tex]
We have to find the area enclose by it.
[tex]x/a=cos\theta, y/b=sin\theta[/tex]
[tex]sin^2\theta+cos^2\theta=x^2/a^2+y^2/b^2[/tex]
Using the formula
[tex]sin^2x+cos^2x=1[/tex]
[tex]\frac{x^2}{a^2}+\frac{y^2}{b^2}=1[/tex]
This is the equation of ellipse.
Area of ellipse
=[tex]4\int_{0}^{a}\frac{b}{a}\sqrt{a^2-x^2}dx[/tex]
When x=0,[tex]\theta=\pi/2[/tex]
When x=a, [tex]\theta=0[/tex]
Using the formula
Area of ellipse
=[tex]\frac{4b}{a}\int_{\pi/2}^{0}\sqrt{a^2-a^2cos^2\theta}(-asin\theta)d\theta[/tex]
Area of ellipse=[tex]-4ba\int_{\pi/2}^{0}\sqrt{1-cos^2\theta}(sin\theta)d\theta[/tex]
Area of ellipse=[tex]-4ba\int_{\pi/2}^{0} sin^2\theta d\theta[/tex]
Area of ellipse=[tex]-2ba\int_{\pi/2}^{0}(2sin^2\theta)d\theta[/tex]
Area of ellipse=[tex]-2ba\int_{\pi/2}^{0}(1-cos2\theta)d\theta[/tex]
Using the formula
[tex]1-cos2\theta=2sin^2\theta[/tex]
Area of ellipse=[tex]-2ba[\theta-1/2sin(2\theta)]^{0}_{\pi/2}[/tex]
Area of ellipse[tex]=-2ba(-\pi/2-0)[/tex]
Area of ellipse=[tex]\pi ab[/tex]
