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?
 A: $$\dot x = 3x + 2y, \quad \dot y = -5x - 3y.$$
$$2\dot x+\dot y = x + y, \quad \dot y+\dot x = -2x - y.$$
Substitute $u=x+y$ and $v=2x+y$:
$$\dot v= u ,\quad \dot u=-v$$
$$\implies \dfrac {dv}{du}=-\dfrac u v$$
$$v^2+u^2=C$$
$$(x+y)^2+(2x+y)^2=C$$
A: late to the party. Given
$$
\left(
\begin{array}{c}
\dot x \\
\dot y
\end{array}
\right) =
\left(
\begin{array}{cc}
p & q \\
r&s
\end{array}
\right)
\left(
\begin{array}{c}
 x \\
 y
\end{array}
\right),
$$
we can have a constant quadratic form (other than at the origin)  only when the trace or the determinant of the coefficient matrix is zero.  Zero determinant  means a line.   More interesting is trace zero.
When
$$
\left(
\begin{array}{c}
\dot x \\
\dot y
\end{array}
\right) =
\left(
\begin{array}{cc}
p & q \\
r&-p
\end{array}
\right)
\left(
\begin{array}{c}
 x \\
 y
\end{array}
\right)
$$
we can readily calculate that
$$ \frac{d}{dt} \; \left( rx^2 - 2pxy - q y^2 \right)   = 0 $$
The type of conic section we see is determined by the discriminant
$$  \Delta = 4p^2 + 4qr  ,$$  as well as the sign of the constant $C$ in $ rx^2 - 2pxy - q y^2   = C $
A: If you set $\vec\alpha=(C,E)$ and $\vec\beta=(D,F)$, then the parametric equation can be written as $\vec p(t)=\vec\alpha\cos t+\vec\beta\sin t$ and vectors $\vec\alpha$, $\vec\beta$ are two conjugate semi-diameters.
But conjugate semidiameters are related to semi-axes $a$, $b$ of the ellipse by Apollonius' formulas:
$$
a^2+b^2=|\vec\alpha|^2+|\vec\beta|^2,
\quad
ab=|\vec\alpha\times\vec\beta|
$$
and can thus be computed. To find the directions of the axes one could follow a nice geometric construction, or find the values of $t$ for which the derivative of $|\vec p(t)|^2$ vanishes, obtaining:
$$
\tan 2t=-{2\vec\alpha\cdot\vec\beta
\over|\vec\alpha|^2+|\vec\beta|^2}.
$$
A: 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.
