lissajous curve ellipse derivation It is known that if $x$ and $y$ are oscillating with the same frequency and a difference in phase, then the curve they trace is generally an ellipse.
I tried to show this but failed. Can you help me with this? Or please tell me where I can find the derivations.
General Formula:[
\begin{cases}
x=A\cos(\theta+\theta_0)\\
y=\sin(\theta)
\end{cases}
]

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.
