In this rotating ellipse formula: $A(x − h)^2 + B(x − h)(y − k) + C(y − k)^2 = 1$
Suppose I have $A,B,C,h,k$ parameters, and I want to obtain the angle $θ$ from the centroid $(h,k)$ to the horizontal axis, at the red dot in this image:
There are two things I am concerning:
Acquiring $θ$: I got the inverse tangent in my mind. If I can get the $(x',y')$ of the red dot, then this should be simple. However, before doing that, I am curious about the fastest way possible to calculate $θ$ in this case. If we have $A,B,C,h,k,$ what would be the fastest way to obtain $θ$?
Due to the nature of an ellipse, both ends of each axis are the same. This implies $0$ degrees and $180$ degrees will look exactly the same visually, however, if possible, I would like to obtain this difference as well. Currently I do not know whether from the rotating formula above, can we obtain the whole $360$ degrees orientation, or only $0-180$? Can somebody please clear this?
$h$ and $k$ don't matter. It looks like the red dot is on the major axis. So this is the angle $\theta$ of my answer to your last question. If $B \ne 0$,
$$ \theta = \arctan \left(\dfrac{C-A + \sqrt{(C-A)^2 + B^2}}{B} \right)$$
I'm not sure what you mean by your second question. What difference do you want to obtain if there is no difference? The formula above gives the $\theta$ that is in the interval $(-\pi/2, \pi/2)$ in radians, or $(-90, 90)$ in degrees.