# Easy ellipse question

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?

The parametric form of an arbitrary ellipse is

\begin{align*} x &= \alpha\cos t + c_x\\ y &= \beta\sin t + c_y \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?

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...

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
\tag{1}\begin{align*} x &=r\cos t\\ y &=r\sin t. \end{align*}​

$$(x+1)^2 + 4y^2 = 4$$​
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} x &= f(u,v)\\ y &= g(u,v) \end{aligned}​
(for new variables $$u$$ and $$v$$), then
$$$$\tag{2}\frac{u^2}{r^2} + \frac{v^2}{r^2} = 1.$$$$​
Once we do so, we can put (2) in parametric form as
\tag{3}\begin{aligned} u &= r\cos t\\ v &= r\sin t \end{aligned}​
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
\tag{4}\begin{aligned} x &= 2u-1\\ y &= v. \end{aligned}​
Plugging the inverse substitution
\begin{aligned} u &= \frac{x+1}{2}\\ v &= y \end{aligned}​
(and r=1) into (3), we have
\begin{align*} \frac{x+1}{2} &= \cos t\\ y &= \sin t \end{align*}​