How to get the correct angle of the ellipse after approximation I need to get the correct angle of rotation of the ellipses. These ellipses are examples. I have a canonical coefficients of the equation of the five points.
$$Ax ^ 2 + Bxy + Cy ^ 2 + Dx + Ey + F = 0$$
Ellipses:

Points: 
Zero ellipse:   [16,46]  [44,19]  [50,35]  [31,61]  [17,54]
First ellipse:  [14,95]  [47,71]  [55,83]  [23,107] [16,103]
Second ellipse: [12,128] [36,117] [58,128] [35,146] [13,136]
Third ellipse:  [16,164] [29,157] [54,188] [40,195] [17,172]
Fourth ellipse: [22,236] [31,207] [50,240] [40,252] [26,244]

Coefficients:
Zero ellipse                 First ellipse                Second ellipse                Third ellipse                  Fourth ellipse A: There're two principal axes in general, so
\begin{align*}
  \theta &=\frac{1}{2} \tan^{-1} \frac{B}{A-C}+\frac{n\pi}{2} \\
  &= \tan^{-1}
     \left(
       \frac{C-A}{B} \color{red}{\pm} \frac{\sqrt{(A-C)^{2}+B^{2}}}{B} \:
     \right) \\
\end{align*}
The centre is given by $$(h,k)=
\left(
  \frac{2CD-BE}{B^2-4AC}, \frac{2AE-BD}{B^2-4AC}
\right)$$
Transforming to
$$\frac{A+C \color{red}{\pm} \sqrt{(A-C)^{2}+B^{2}}}{2} x'^2+
\frac{A+C \color{red}{\mp} \sqrt{(A-C)^{2}+B^{2}}}{2} y'^2+
\frac
{\det
  \begin{pmatrix}
    A & \frac{B}{2} & \frac{D}{2} \\
    \frac{B}{2} & C & \frac{E}{2} \\
    \frac{D}{2} & \frac{E}{2} & F
  \end{pmatrix}}
{\det
  \begin{pmatrix}
    A & \frac{B}{2} \\
    \frac{B}{2} & C \\
  \end{pmatrix}}=0$$
where $\begin{pmatrix} x' \\ y' \end{pmatrix}=
\begin{pmatrix}
   \cos \theta & \sin \theta \\
  -\sin \theta & \cos \theta
\end{pmatrix}
\begin{pmatrix} x-h \\ y-k \end{pmatrix}$.
The axes will match, up to reflection about the axes of symmetry, when the $\color{red}{\text{case}}$ (upper or lower) agree.

Numerical example
Given five points: $(2,1)$, $(1,1)$, $(-2,-2)$, $(-1,-2)$, $(1,-1)$
$A=1$, $B=-2$, $C=2$, $D=-1$, $E=2$, $F=-2$
$$(h,k)=(0,-0.5)$$
$$\det
  \begin{pmatrix}
    A & \frac{B}{2} & \frac{D}{2} \\
    \frac{B}{2} & C & \frac{E}{2} \\
    \frac{D}{2} & \frac{E}{2} & F
  \end{pmatrix} = ACF-\frac{A E^2+C D^2+F B^2-EDB}{4}=-\frac{5}{2}$$
$$\det
  \begin{pmatrix}
    A & \frac{B}{2} \\
    \frac{B}{2} & C
  \end{pmatrix} = -\frac{B^2}{4}+AC=1$$
$$\frac{A+C \pm \sqrt{(A-C)^{2}+B^{2}}}{2}=\frac{3 \pm \sqrt{5}}{2}$$
Using upper case convention:
$$\frac{3+\sqrt{5}}{2} x'^2+\frac{3-\sqrt{5}}{2} y'^2=\frac{5}{2}$$
$$\frac{x'^2}{a^2}+\frac{y'^2}{b^2}=1$$
$$(x',y')= (a\cos t,b\sin t)$$
where $\displaystyle \begin{pmatrix} a \\ b \end{pmatrix}=
\begin{pmatrix}
  \sqrt{\frac{5}{3+\sqrt{5}}} \\
  \sqrt{\frac{5}{3-\sqrt{5}}}
\end{pmatrix}$
$$\theta = \tan^{-1}
     \left(
       \frac{C-A}{B}+\frac{\sqrt{(A-C)^{2}+B^{2}}}{B} \:
     \right) =
  \tan^{-1} \left( -\frac{\sqrt{5}+1}{2} \right)
  \approx -58.28^{\circ} $$
\begin{align*}
    \begin{pmatrix} x \\ y \end{pmatrix} &=
    \begin{pmatrix}
      \cos \theta & -\sin \theta \\
      \sin \theta &  \cos \theta
    \end{pmatrix}
    \begin{pmatrix} x' \\ y' \end{pmatrix}+
    \begin{pmatrix} h \\ k \end{pmatrix} \\  &&\\
    &=
    \begin{pmatrix} 
      h+x'\cos \theta-y'\sin \theta \\
      k+x'\sin \theta+y'\cos \theta \end{pmatrix} \\  &&\\
    &=
    \begin{pmatrix} 
      \sqrt{\frac{5}{2}+\sqrt{5}\,} \, \sin t+
      \sqrt{\frac{5}{2}-\sqrt{5}\,} \, \cos t \\
      -\frac{1}{2}+
      \frac{\sqrt{5+\sqrt{5}}}{2} \, \sin t-
      \frac{\sqrt{5-\sqrt{5}}}{2} \, \cos t \end{pmatrix}
  \end{align*}

A: Justification of the formula:
After centering (translation to let the linear terms vanish), the equation becomes
$$Ax^2+Bxy+Cy^2+F'=0.$$
Then you apply a rotation of angle $\theta$ to let the mixed term $Bxy$ vanish from the quadratic terms,
$$A(x\cos\theta-y\sin\theta)^2+B(x\cos\theta-y\sin\theta)(x\sin\theta+y\cos\theta)+C(x\sin\theta+y\cos\theta)^2.$$
This is achieved when
$$2(C-A)\cos\theta\sin\theta+2B(\cos^2\theta-\sin^2\theta)=(C-A)\sin(2\theta)+2B\cos(2\theta)=0.$$
(Then you can rewrite $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$.)
Hence,
$$\tan(2\theta)=\frac{2B}{A-C}.$$
The angle $\theta$ is undeterminate by a multiple of a quater turn, as the ellipse has two symmetry axis.
A: Although it sounds like a question, for calculation did you use atan2 function or atan function? Quadrant placement is also important.
