I hava an ellipse. I know its' equation.
For a given 't' how can I calculate angle "alpha". (I also know the coordinate of the tangent point)
[Math] How to calculate the tangent angle with the axis of an ellipse
calculus
Related Solutions
Let $X$ be the external point. Let $P$ be the center of the circle. Let $Q$ be a point of tangency. Then triangle $\Delta PQX$ is a right triangle. If we let $\theta$ be the angle that $PX$ makes with the horizontal, then the angle of the tangent point with the horizontal is $\theta + \angle XPQ$ and $\theta - \angle XPQ$ respectively. Consequently, the angle that the tangents make with the vertical will also be $\theta \pm \angle XPQ$.
In the figure above, the green angle would be the angle give by $\theta + \angle XPQ$.
If $X = (x_2, y_2)$ and $P = (x_1, y_1)$ then $\theta = \arctan\left(\frac{y_2 - y_1}{x_2 - x_1}\right)$.
If you want the heading to the tangent point, then you will want the red angle (there is another angle for the lower tangent which I have not labelled, it is given analogously) which will be $90^{\circ} - (\theta - \angle XPQ)$.
If you want the angle the tangent makes with the vertical, you will want the blue angle (or the supplement of the blue angle) which is equal to the green angle, $\theta + \angle XPQ$.
You may have to tweak a bit to make sure that everything works out in all quadrants, but this is the basic idea.
Edit If you want $\angle FXQ$ as defined in your comment, then it is simple. Draw a vertical line intersecting $X$. Then the tangent cuts through the vertical line and the $y$ axis as parallel lines. $\angle FXQ$ is then given by the blue angle.
I just wrote a new answer to the question you referenced. That post describes how you can get from true anomaly $\theta$ via eccentric anomaly $E$ to mean anomaly $M$ which is proportional to both area and time.
To find the change in true anomaly $\theta$ corresponding to a given change in area, you'd first turn the original angle $\theta_1$ into some mean anomaly $M_1$, then you'd add to that and try to reverse the process, i.e. compute $\theta_2$ from $M_2$. But the relation between $M$ and $E$ is Kepler's equation:
$$M=E-e\sin E$$
This is a transcendental equation. Wikipedia writes:
This equation does not have a closed-form solution for $E$ given $M$. It is usually solved by numerical methods, e.g. Newton-Raphson method.
The lemma on Kepler's equation has a whole section on this inverse problem, followed by one about numeric approximations. So I guess that would be the way to go.
Best Answer
The tangent is in the direction of $$\left(\frac{\mathrm{d}x}{\mathrm{d}t},\frac{\mathrm{d}y}{\mathrm{d}t}\right)$$ Then the slope of the tangent is given by $$\frac{\frac{\mathrm{d}y}{\mathrm{d}t}}{\frac{\mathrm{d}x}{\mathrm{d}t}}=\frac{10\cos{t}}{-30\sin{t}}=-\frac{1}{3}\cot{t}$$
The angle is the $\arctan$ of the slope