It seems worthwhile to review how a generally polarized field phasor leads to linear, circular, and elliptic geometries.

The most general field polarized in the $$x, y$$ plane has the form

\label{eqn:polarizationReview:20}
\BE
= \lr{ \xcap a e^{j \alpha} + \ycap b e^{j \beta} } e^{j \lr{ \omega t -k z }}
= \lr{ \xcap a e^{j \lr{\alpha – \beta}/2} + \ycap b e^{j \lr{ \beta – \alpha}/2} } e^{j \lr{ \omega t -k z + \lr{\alpha + \beta}/2 }}.

Knowing to factor out the average phase angle above is only because I tried initially without that and things got ugly and messy. I guessed this would help (it does).

Let $$\boldsymbol{\mathcal{E}} = \text{Re} \BE = \xcap x + \ycap y$$, $$\theta = \omega t + (\alpha + \beta)/2$$, and $$\phi = (\alpha – \beta)/2$$, so that

\label{eqn:polarizationReview:40}
\BE
= \lr{ \xcap a e^{j \phi} + \ycap b e^{-j \phi} } e^{j \theta }.

The coordinates can now be read off

\label{eqn:polarizationReview:60}
\frac{x}{a} = \cos\phi \cos\theta – \sin\phi \sin\theta

\label{eqn:polarizationReview:80}
\frac{y}{b} = \cos\phi \cos\theta + \sin\phi \sin\theta,

or in matrix form

\label{eqn:polarizationReview:100}
\begin{bmatrix}
x/a \\
y/b \\
\end{bmatrix}
=
\begin{bmatrix}
\cos\phi & – \sin\phi \\
\cos\phi & \sin\phi
\end{bmatrix}
\begin{bmatrix}
\cos\theta \\
\sin\theta
\end{bmatrix}.

The goal is to eliminate all the $$\theta$$ (i.e. time dependence), converting the parametric relationship into a conic form.
Assuming that neither $$\cos\theta$$, nor $$\sin\theta$$ are zero for now (those are special cases and lead to linear polarization), inverting the matrix will allow the $$\theta$$ dependence to be eliminated

\label{eqn:polarizationReview:120}
\inv{\sin\lr{ 2\phi }}
\begin{bmatrix}
\sin\phi & \sin\phi \\
– \cos\phi & \cos\phi
\end{bmatrix}
\begin{bmatrix}
x/a \\
y/b \\
\end{bmatrix}
=
\begin{bmatrix}
\cos\theta \\
\sin\theta
\end{bmatrix}.

Squaring and summing both rows of these equation gives
\label{eqn:polarizationReview:140}
\begin{aligned}
1
&=
\inv{\sin^2 \lr{ 2\phi}}
\lr{
\sin^2\phi
\lr{
\frac{x}{a}
+\frac{y}{b}
}^2
+
\cos^2\phi
\lr{
-\frac{x}{a}
+\frac{y}{b}
}^2
} \\
&=
\inv{\sin^2 \lr{ 2\phi}}
\lr{
\frac{x^2}{a^2}
+\frac{y^2}{b^2}
+2 \frac{x y}{a b} \lr{ \sin^2\phi – \cos^2\phi }
} \\
&=
\inv{\sin^2 \lr{ 2\phi}}
\lr{
\frac{x^2}{a^2}
+\frac{y^2}{b^2}
-2 \frac{x y}{a b} \cos \lr{2\phi}
}
\end{aligned}

Time to summarize and handle the special cases.

1. To have $$\cos\phi = 0$$, the phase angles must satisfy $$\alpha – \beta = \lr{ 1 + 2 k } \pi, \, k \in \mathbb{Z}$$.

For this case \ref{eqn:polarizationReview:50} reduces to

\label{eqn:polarizationReview:160}
-\frac{x}{a} = \frac{y}{b},

which is just a line.

### Example.

Let $$\alpha = 0, \beta = -\pi$$, so that the phasor has the value

\label{eqn:polarizationReview:260}
\BE = \lr{ \xcap a – \ycap b } e^{j \omega t}

2. For have $$\sin\phi = 0$$, the phase angles must satisfy $$\alpha – \beta = 2 \pi k, \, k \in \mathbb{Z}$$.

For this case \ref{eqn:polarizationReview:60} and \ref{eqn:polarizationReview:80} reduce to

\label{eqn:polarizationReview:180}
\frac{x}{a} = \frac{y}{b},

also just a line.

### Example.

Let $$\alpha = \beta = 0$$, so that the phasor has the value

\label{eqn:polarizationReview:280}
\BE = \lr{ \xcap a + \ycap b } e^{j \omega t}

3. Last is the circular and elliptically polarized case. The system is clearly elliptically polarized if $$\cos(2 \phi) = 0$$, or $$\alpha – \beta = (\pi/2)( 1 + 2 k ), k \in \mathbb{Z}$$. When that is the case and $$a = b$$ also holds, the ellipse is a circle.

When the $$\cos( 2 \phi) = 0$$ condition does not hold, a rotation of coordinates

\label{eqn:polarizationReview:200}
\begin{bmatrix}
x \\
y
\end{bmatrix}
=
\begin{bmatrix}
\cos\mu & \sin\mu \\
-\sin\mu & \cos\mu
\end{bmatrix}
\begin{bmatrix}
u \\
v
\end{bmatrix}

where

\label{eqn:polarizationReview:220}
\mu = \inv{2} \tan^{-1} \lr{ \frac{ 2 \cos (\alpha – \beta)}{b – a}}

puts the trajectory into a standard (but messy) conic form

\label{eqn:polarizationReview:240}
1 = \frac{u^2}{ab} \lr{
\frac{b}{a} \cos^2 \mu
+ \frac{a}{b} \sin^2 \mu
+ \inv{2} \sin\lr{2 \mu + \alpha – \beta}
}
+
\frac{v^2}{ab} \lr{
\frac{b}{a} \sin^2 \mu
+ \frac{a}{b} \cos^2 \mu
– \inv{2} \sin\lr{2 \mu + \alpha – \beta}
}

It isn’t obvious to me that the factors of the $$u^2, v^2$$ terms are necessarily positive, which is required for the conic to be an ellipse and not a hyperbola.

### Circular polarization example.

With $$a = b = E_0$$, $$\alpha = 0$$, $$\beta = \pm \pi/2$$, all the circular polarization conditions are met, leaving the phasor with values

\label{eqn:polarizationReview:300}
\BE = E_0 \lr{ \xcap \pm j \ycap } e^{j \omega t}