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}<br />
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*}
