# Thread: parametric equations of an ellipse

1. ## parametric equations of an ellipse

Use the parametric equations of an ellipse
x=9cos
y=3sin
02

to find the area that it encloses

does anyone know this one?

2. Originally Posted by dat1611
Use the parametric equations of an ellipse
x=9cos
y=3sin
02

to find the area that it encloses

does anyone know this one?
1. The formula to calculate the area of an ellipse (in standard form) is:

$A=\pi \cdot a \cdot b$

2. Eliminate the parameter $\theta$

$\left|\begin{array}{rcl}x&=&9\cos(\theta) \\ y &=& 3\sin(\theta)\end{array}\right.$ $\implies$ $\left|\begin{array}{rcl}\dfrac{x^2}{81} &=&\cos^2(\theta) \\ \dfrac{y^2}9 &=&\sin^2(\theta)\end{array}\right.$

$\dfrac{x^2}{9^2}+\dfrac{y^2}{3^2}=1$

3. Now use the formula from #1.

3. whats a and whats b

4. Originally Posted by dat1611
whats a and whats b
a and b are the major and minor radii of the ellipse. You can find these easily from your parametric equations by finding the maximum values for x and y.
This is so because your ellipse is simply a stretching of a unit circle (x = cos t, y = sin t) along the axes. No rotation is involved.

5. Originally Posted by dat1611
Use the parametric equations of an ellipse
x=9cos
y=3sin
02

to find the area that it encloses

does anyone know this one?
Coordinate axes coincide with symmetry axes of the ellipse (see picture), ie сoordinate axes divide the ellipse into four equal parts. The fourth part of the required area $A$, which is located in the first quadrant, we will find as the area of a curvilinear trapezoid, that adjacent to the X-axis: $\frac{1}{4}A = \int\limits_0^9 {ydx} .$
Now we transform the integral to the variable $\theta$, using ellipse parametric equations:

$y = 3\sin \theta ;{\text{ }}x = 9\cos \theta \Rightarrow dx = - 9\sin \theta d\theta.$

Find the lower integral limit:

If $x = 0$ then $9\cos \theta = 0 \Rightarrow \theta = \frac{\pi }{2}.$

Find the upper integral limit [we take the value x = 9, because the maximum value of cosine is a unit]:

If $x = 9$ then $9\cos \theta = 9 \Rightarrow \theta = 0.$

So we have

$\frac{1}{4}A = \int\limits_0^9 {ydx} = \int\limits_{\pi /2}^0 {3\sin \theta \left( { - 9\sin \theta } \right)d\theta } = - 27\int\limits_{\pi /2}^0 {{{\sin }^2}\theta d\theta } =$

$= \frac{{27}}{2}\int\limits_0^{\pi /2} {\left( {1 - \cos 2\theta } \right)d\theta } = \frac{{27}}{2}\left. {\left( {\theta - \frac{1}{2}\sin 2\theta } \right)} \right|_0^{\pi /2} = \frac{{27}}{4}\pi .$

Finally we have

$\frac{1}{4}A = \frac{{27}}{4}\pi \Leftrightarrow \boxed{A = 27\pi }$

See this picture