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:

1. 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?

2. 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?

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:

1. 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 $θ$?

2. 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?