conic sectionsparametrization
Given an ellipse of semi-axes $a$ and $b$ centered on the point $(x_0,y_0)$,
$$\frac{(x-x_0)^{2}}{a^{2}}+\frac{(y-y_0)^{2}}{b^{2}}=1$$
It can be expressed using parametric coordinates:
$$\vec{F}(t)=(x(t), y(t))$$
$$x(t)=a \cos (t)+x_0 $$
$$ y(t)=b \sin (t)+y_0$$
Then, if $P(x(t),y(t))$ is a point belonging to the ellipse, how can we know whether it will move clockwise or counterclockwise when the parameter $t$ is varied?
The $\theta$ represents the angle between the $x$-axis and the line which passes the center and $(x_1,y_1)$.
In other words, the $\theta$ satisfies $$\tan\theta=\frac{y_1}{x_1}$$ if $\theta\not=\frac{\pi}{2}.$
See this figure.
I'm polishing this post on Stack Overflow with proper math formatting and a nicer example.
The following description follows the German Wikipedia article Hauptachsentransformation. Its English counterpart, according to inter-wiki links, is principal component analysis. I find the former article a lot more geometric than the latter. The latter has a strong focus on statistical data, which might be useful for some applications.
Your ellipse is described as
$$(x, y)\cdot\begin{pmatrix}A&C/2\\C/2&B\end{pmatrix}\cdot\begin{pmatrix}x\\y\end{pmatrix}+(D,E)\cdot\begin{pmatrix}x\\y\end{pmatrix}+F=0\tag1\\ (x, y)\cdot\left(\begin{array}{rr}4.36&-2.52\\-2.52&2.89\end{array}\right)\cdot\begin{pmatrix}x\\y\end{pmatrix}+(30.8,-0.6)\cdot\begin{pmatrix}x\\y\end{pmatrix}+81=0$$
First you identify the rotation.
You do this by identifying the eigenvalues and eigenvectors of this $2\times2$ matrix. These eigenvectors will, after normalization, form an orthogonal matrix describing the rotation. I'll not go into the details of how to find eigenvectors of a matrix.
Using the example, you find that the eigenvalues of the matrix are $\lambda_1=1$ and $\lambda_2=6.25$. (Taking the eigenvalue with smaller absolute value first will give you the semimajor axis first in the final result.) For the former, an eigenvector is $(3,4)^T$, for the latter $(-4,3)^T$ is one. Normalizing them to unit length yields $(0.6,0.8)^T$ resp. $(-0.8,0.6)^T$. So the diagonal decomposition is
$$\left(\begin{array}{rr}4.36&-2.52\\-2.52&2.89\end{array}\right)= \left(\begin{array}{rr}0.6&-0.8\\0.8&0.6\end{array}\right)\cdot \left(\begin{array}{ll}1&0\\0&6.25\end{array}\right)\cdot \left(\begin{array}{rr}0.6&0.8\\-0.8&0.6\end{array}\right)\tag2$$
The first of the three factors has the eigenvectors as columns, the last one is the transpose of the first and therefore has them as rows. The diagonal matrix in the center has the eigenvalues, in matching order.
If you want to avoid computing eigenvalues and eigenvectors, you can also use the fact that you know what a rotation matrix looks like, and start with the equation
$$\begin{pmatrix}\cos\theta&\sin\theta\\-\sin\theta&\cos\theta\end{pmatrix}\cdot\begin{pmatrix}A&C/2\\C/2&B\end{pmatrix}\cdot\begin{pmatrix}\cos\theta&-\sin\theta\\\sin\theta&\cos\theta\end{pmatrix}\overset!=\begin{pmatrix}\lambda_1&0\\0&\lambda_2\end{pmatrix}$$
Now take one of the entries which has to be zero, and perform the matrix multiplication to obtain the corresponding left hand side. You will get
\begin{align*} (B-A)\sin\theta\cos\theta &+ \tfrac12C\bigl(\cos^2\theta-\sin^2\theta\bigr) = 0 \tag3 \\ 2(B-A)\sin\theta\cos\theta &= C\bigl(\sin^2\theta-\cos^2\theta\bigr) \\ 4(B-A)^2\sin^2\theta\cos^2\theta &= C^2\Bigl(\bigl(\sin^2\theta-\cos^2\theta\bigr)^2 - \bigl(\sin^2\theta+\cos^2\theta\bigr)^2 + 1\Bigr) \\ 4(B-A)^2\sin^2\theta\cos^2\theta &= C^2\Bigl(1-4\sin^2\theta\cos^2\theta\Bigr) \\ u := 4\sin^2\theta\cos^2\theta &= \frac{C^2}{(B-A)^2+C^2} \\ 4\sin^2\theta\bigl(1-\sin^2\theta\bigr) &= u \\ \sin^4\theta - \sin^2\theta + \tfrac14u &= 0 \\ \sin^2\theta &= \frac{1\pm\sqrt{1-u}}{2} \end{align*}
So for our example this gives us
$$ u = \frac{C^2}{(B-A)^2+C^2} = \frac{5.04^2}{(2.89 - 4.36)^2+5.04^2} = 0.9216 $$
\begin{align*} \sin^2\theta_1 &= \frac{1-\sqrt{0.0784}}2 = 0.36 & \lvert\sin\theta_1\rvert &= \sqrt{0.36} = 0.6 & \lvert\cos\theta_1\rvert &= \sqrt{1-0.36} = 0.8 \\ \sin^2\theta_2 &= \frac{1+\sqrt{0.0784}}2 = 0.64 & \lvert\sin\theta_2\rvert &= \sqrt{0.64} = 0.8 & \lvert\cos\theta_2\rvert &= \sqrt{1-0.64} = 0.6 \end{align*}
At this moment there appear to be $8$ solutions: $\theta_1$ and $\theta_2$, each with $2\cdot2=4$ possible sign combinations. But half of these will not satisfy equation $(3)$. The other half corresponds to rotations which differ by multiples of $90°$. So choosing between $\theta_1$ and $\theta_2$ you essentially choose which axis of the ellipse gets aligned with which axis of the coordinate system, and by choosing among the remaining sign combinations you can rotate things by $180°$ which does not change the ellipse.
For the sake of consistency with what I wrote above, I'l choose $\sin\theta=0.8$ and $\cos\theta=0.6$. Then the diagonal matrix can be computed (equivalent to $(2)$ above) as
$$\left(\begin{array}{rr}0.6&0.8\\-0.8&0.6\end{array}\right)\cdot \left(\begin{array}{rr}4.36&-2.52\\-2.52&2.89\end{array}\right)\cdot \left(\begin{array}{rr}0.6&-0.8\\0.8&0.6\end{array}\right)= \left(\begin{array}{ll}1&0\\0&6.25\end{array}\right)$$
If you multiply the vector $(x,y)^T$ with the last matrix from $(2)$, then you will change the coordinate system in such a way that the mixed term vanishes, i.e. the $x$ and $y$ axes are parallel to the main axes of your ellipse. I'll write coordinates in that system as $(x',y')^T$:
$$\begin{pmatrix}x'\\y'\end{pmatrix}=\left(\begin{array}{rr}0.6&0.8\\-0.8&0.6\end{array}\right)\cdot\begin{pmatrix}x\\y\end{pmatrix}\tag4$$
The rotation of the ellipse can be read from that rotation matrix. The matrix used in $(3)$ transforms a point on the rotated ellipse into a point on the axis-aligned ellipse. So the direction is opposite to what you'd use when describing the rotation of the ellipse, and you best compute the angle from the first row of that matrix:
$$\theta=\arctan\frac{\sin\theta}{\cos\theta}=\arctan\frac{0.8}{0.6}\approx53.13°$$
If you multiply the row vector $(D,E)$ in equation $(1)$ with the first of the three matrices from $(2)$, then this effect will exactly cancel the multiplication expressed in $(4)$. Therefore in that changed coordinate system, you use the central diagonal matrix for the quadratic term, and this product for the linear term.
$$(30.8,-0.6)\cdot\left(\begin{array}{rr}0.6&-0.8\\0.8&0.6\end{array}\right)=(18,-25)\\ (x', y')\cdot\left(\begin{array}{ll}1&0\\0&6.25\end{array}\right)\cdot\begin{pmatrix}x'\\y'\end{pmatrix}+(18,-25)\cdot\begin{pmatrix}x'\\y'\end{pmatrix}+81=0\\ (1\,x'^2 + 18\,x')+(6.25\,y'^2-25\,y')+81=0$$
Now you have to complete the square independently for $x'$ and $y'$, and you end up with a form from which you can read the center coordinates.
\begin{align*} x'^2 + 18\,x' &= \phantom{1.00}(x' + 9)^2 - 81 & x_c' &= -9 \\ 6.25\,y'^2 - 25\,y' &= 6.25(y' - 2)^2 - 25 & y_c' &= \phantom+2 \\ \end{align*}
Note that the center coordinates $x_c'$ and $y_c'$ describe the center in the already rotated coordinate system. To obtain the original center you'd multiply again with the first matrix from $(2)$:
$$\begin{pmatrix}x_c\\y_c\end{pmatrix}= \left(\begin{array}{rr}0.6&-0.8\\0.8&0.6\end{array}\right)\cdot \left(\begin{array}{r}-9\\2\end{array}\right)= \begin{pmatrix}-7\\-6\end{pmatrix}$$
You can also do a second change of coodinate system, this time into one where the ellipse is not only axis-aligned but also centered at the origin.
The completed squares above contributed some more terms to the constant factor $F$, since the equation by now has become
$$(x' + 9)^2 - 81 + 6.25(y' - 2)^2 - 25 + 81 = 0$$
The combined constant term here is $81 - 81 - 25 = -25$. Now you move this to the right side of the equation, then divide the whole equation by this number so that you get the $=1$ from the desired form. Then you can deduce the radii $a$ and $b$.
\begin{align*} (x' + 9)^2 + 6.25(y' - 2)^2 &= 25 \\ \frac{(x' + 9)^2}{25} + \frac{6.25(y' - 2)^2}{25} &= 1 \end{align*}
\begin{align*} \frac1{a^2}&=\frac{1}{25} & a &= \sqrt{\frac{25}{1}}\phantom{.0} = 5 \\ \frac1{b^2}&=\frac{6.25}{25} & b &= \sqrt{\frac{25}{6.25}} = 2 \end{align*}
Now let's check that we didn't make any mistakes. With the parameters we found, you can piece together the equation
\begin{align*} \left(\frac{0.6(x-(-7))+0.8(y-(-6))}{5}\right)^2+ \left(\frac{0.6(y-(-6))-0.8(x-(-7))}{2}\right)^2 &= 1 \\ \frac{\bigl(0.6(x+7)+0.8(y+6)\bigr)^2}{25}+ \frac{\bigl(0.6(y+6)-0.8(x+7)\bigr)^2}{4} &= 1 \\ 4\bigl(0.6(x+7)+0.8(y+6)\bigr)^2+ 25\bigl(0.6(y+6)-0.8(x+7)\bigr)^2 - 100 &= 0 \\ &\;\vdots \\ 17.44\,x^2 + 11.56\,y^2 - 20.16\,xy + 123.2\,x - 2.4\,y + 324 &= 0 \\ 4.36\,x^2 + 2.89\,y^2 - 5.04\,xy + 30.8\,x - 0.6\,y + 81 &= 0 \end{align*}
So this is indeed the right equation, equivalent to the one given in the original example.
With a few sign changes, the above works for hyperbolas as well.
The approach using eigenvectors will generalize to ellipsoids in space. The approach using the form of the rotation matrix is harder in space.
Best Answer
Assuming that $a,b>0$, the movement will be counterclockwise. Note that $F'(0)=(0,b)$, which points upwards. So, since $F(0)$ is the point of the ellipse located more to the right, it starts moving counterclockwise. Since $F'(t)$ is never $0$, the movement cannot become clockwise after.