Understanding the Parametric Form of an Ellipse: Step-by-Step Solution

[winbacker]

I have the solution to a problem and I need help understanding how the solution was obtained.

I have to take an equation of an ellipse and transfer it to parametric form.

The ellipse has the equation: (x+1)^2 + 4y^2 = 4.

The solution has the ellipse parameterized as follows:

x+1 = 2cost and y = sint

Can someone just explain this translation, step by step?

 

[foxjwill]

The parametric form of an arbitrary ellipse is

$$\begin{align*}<br /> x &= \alpha\cos t + c_x\\ <br /> y &= \beta\sin t + c_y<br /> \end{align*}$$

where $$(c_x, c_y)$$ is its center and $$\alpha,\beta$$ are constants.

How would you use this to convert your equation to parametric form? Can you find values for any of $$c_x,c_y,\alpha$$, or $$\beta$$ right off the bat?

 

[winbacker]

well, I assume the center of the ellipse is at (-1,0). Otherwise I am stumped...how do they get 2cost?? And how is y = sint? I am very confused...

 

[foxjwill]

You know what, I just thought of a better way to do this.

Recall that in parametric form the equation $$\frac{x^2}{r^2}+\frac{y^2}{r^2}=1$$ is

$$\begin{equation}\tag{1}\begin{align*}<br /> x &=r\cos t\\<br /> y &=r\sin t.<br /> \end{align*}\end{equation}$$​

So, to put your equation

$$(x+1)^2 + 4y^2 = 4$$​

into parametric form, we need to first do some sort of substitution of variables to get it to look like the equation of a circle. In other words, we want to find functions $$f$$ and $$g$$ such that if

$$\begin{aligned}<br /> x &= f(u,v)\\<br /> y &= g(u,v)<br /> \end{aligned}$$​

(for new variables $$u$$ and $$v$$), then

$$\begin{equation}\tag{2}\frac{u^2}{r^2} + \frac{v^2}{r^2} = 1.\end{equation}$$​

Once we do so, we can put (2) in parametric form as

$$\begin{equation}\tag{3}\begin{aligned}<br /> u &= r\cos t\\<br /> v &= r\sin t<br /> \end{aligned}\end{equation}$$​

and then (since u will turn out to only depend on x, and v only on y) we can plug in the inverse substitution of (2) to get the parametrization we want.

Let's start by dividing the equation by 4:

$$\frac{(x+1)^2}{4}+y^2=1.$$​

The obvious choice for y is to let $$y=v\ (=g(u,v))$$.

x, though, is not quite as simple. However, a major benefit of our choice for y is that since $$\frac{(x+1)^2}{4}$$ doesn't depend on y, it also doesn't depend on v. Thus, instead of finding a function f(u,v), we just need to find a function f(u)! Additionally, notice that by letting y=u, we have effectively chosen r to be 1. So, to we need to find a function f such that

$$\frac{(f(u)+1)^2}{4}=\frac{u^2}{1}=u^2.$$​

Solving, we get

$$f(u)=2u-1\ (=x).$$​

So, to consolidate, we have

$$\begin{equation}\tag{4}\begin{aligned}<br /> x &= 2u-1\\<br /> y &= v.<br /> \end{aligned}\end{equation}$$​

Plugging the inverse substitution

$$\begin{equation}\begin{aligned}<br /> u &= \frac{x+1}{2}\\<br /> v &= y<br /> \end{aligned}\end{equation}$$​

(and r=1) into (3), we have

$$\begin{align*}<br /> \frac{x+1}{2} &= \cos t\\<br /> y &= \sin t<br /> \end{align*}$$​