lissajous curve ellipse derivation

You can write $\cos(\theta+\theta_0)=\cos(\theta_0)\cos(\theta)-\sin(\theta_0)\sin(\theta)$ so the parametric equations are of the form

$$ \begin{cases} x=a\cos(\theta)+b\sin(\theta)\\ y=c\sin(\theta)+d\sin(\theta) \end{cases} $$

which can be written as a matrix equation:

$$ \begin{bmatrix} x \\ y \end{bmatrix} = \begin{bmatrix} a & b\\ c & d \end{bmatrix} \begin{bmatrix} \cos(\theta) \\ \sin(\theta) \end{bmatrix} $$

or $X=MY$ with $X=\begin{bmatrix} x \\ y \end{bmatrix} $, $M=\begin{bmatrix} a & b \\ c & d \end{bmatrix} $ and $Y=\begin{bmatrix} \cos(\theta) \\ \sin(\theta) \end{bmatrix}$.

If $M$ is invertible (this is the case in the OP, as shown by a quick computation), we can write $Y=M^{-1}X$. We can then transform the identity $\cos^2(\theta)+\sin^2(\theta)=1$ into a quadratic equation in $x$ and $y$. Using the discriminant (for example), you can show this is the equation of an ellipse.

Detail of computations:

More specifically, $M^{-1}=\frac{1}{\delta}\begin{bmatrix} d & -b \\ -c & a \end{bmatrix}$ with $\delta=ad-bc$. Therefore,

$$ \cos(\theta)=\frac{1}{\delta}(dx-by),\hskip 5mm \sin(\theta)=\frac{1}{\delta}(-cx+ay) $$

so

$$ (dx-by)^2+(ay-cx)^2=\delta^2 \iff (a^2+d^2)x^2-2(bd+ac)xy+(b^2+c^2)y^2-\delta^2 = 0 $$

Therefore, the graph is indeed a conic section. It is not degenerate since it is the image of a circle by an invertible linear function (hence cannot be empty, or a point, or one or two lines).

The discriminant is $\Delta=4(bd+ac)^2-4(a^2+d^2)(b^2+c^2)=-4(ab-cd)^2$ is clearly negative, so it is an ellipse.

Remark: it cannot be a hyperbola or a parabola for topological reasons too.