<s>[INST] Sketching an arbitrary ellipse from its parametric equation I was solving a question, in which I was asked to solve the following system of differential equations:
$$
\dot x = 3x + 2y, \quad \dot y = -5x - 3y.
$$
I got the general solution in $\mathbb R^2$ to be
$$
\begin{pmatrix} x \\ y \end{pmatrix} = \begin{pmatrix} C & D \\ -\frac32 C + \frac12 D & -\frac12 C - \frac 32 D \end{pmatrix}\begin{pmatrix} \cos t \\ \sin t \end{pmatrix}, \quad C,D\in \mathbb R.
$$
Now, I want to sketch the phase portrait of this solution (apparently, as we change $C$ and $D$, only the scale of the ellipse changes, not its orientation or ratio between major and minor axes). My question is, how do I determine the orientation of axis lengths of this ellipse (in terms of $C$ and $D$)? Given an arbitrary ellipse of the form
$$
\begin{pmatrix} C & D \\ E & F \end{pmatrix}\begin{pmatrix} \cos t \\ \sin t \end{pmatrix},
$$
how would one sketch it?
 [/INST] First write $\cos t$ and $\sin t$ in terms of $x$ and $y$:
$$
\begin{pmatrix} x \\ y\end{pmatrix}=\begin{pmatrix} C & D \\ E & F \end{pmatrix}\begin{pmatrix} \cos t \\ \sin t \end{pmatrix}\\
\begin{pmatrix} \cos t \\ \sin t \end{pmatrix}=\begin{pmatrix} C & D \\ E & F \end{pmatrix}^{-1}\begin{pmatrix} x \\ y\end{pmatrix}
$$
The inverse is a $2\times2$ matrix, say $\begin{pmatrix} \alpha & \beta \\ \gamma & \delta \end{pmatrix}$.
You also know $\cos^2 t+\sin^2t=1$. So $$(\alpha x+\beta y)^2+(\gamma x+\delta y)^2=1$$
You can group the terms together, then use this answer to get the orientation, and something like this answer to find the length of the semiaxes.
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