Viewing the conic section [math]\phi[/math] maps out on the DC sector plane, we know that it follows an elliptical path centered on it's x axis. Performing a passive rotation on points in the DC section plane does not physically change the position in space, i.e. passive rotations only give the components in a new coordinate system. Once such a rotation has been performed, the equation describing these points must be done within that plane.
An ellipse centered at the origin can be expressed in the form
[math]\frac{x^2}{a^2}+\frac{y^2}{b^2}=1[/math]
For an ellipse not centered on the origin, but instead the point (h',k'), this expression becomes
[math]\frac{(x+h')^2}{a^2}+\frac{(y+k')^2}{b^2}=1[/math]
In the plane of the DC sector, this equation becomes
[math]\frac{(x'+\Delta a)^2}{a^2}+\frac{(y')^2}{b^2}=1[/math]
where the center of the ellipse is found at [math]\{\Delta a, 0\}[/math].
Switching to the frame of the wires, the ellipse is still centered at [math]\{\Delta a,0\}[/math] in the DC sector, with the semi-major axis lying on the x' axis. For a rotation in the y-x plane, this corresponds to a positive angle [math]\theta[/math], with the rotation matrix [math]R(\theta_{yx})[/math]. In the frame of the wires, this center point falls at
[math]\begin{bmatrix}
x'' \\
y'' \\
z''
\end{bmatrix}=
\begin{bmatrix}
cos\ 6^{\circ} & -sin\ 6^{\circ} & 0 \\
sin\ 6^{\circ} &cos\ 6^{\circ} & 0 \\
0 & 0 & 1
\end{bmatrix}\cdot
\begin{bmatrix}
x' \\
y' \\
z'
\end{bmatrix}[/math]
[math]\begin{bmatrix}
x'' \\
y'' \\
z''
\end{bmatrix}=
\begin{bmatrix}
cos\ 6^{\circ} & -sin\ 6^{\circ} & 0 \\
sin\ 6^{\circ} &cos\ 6^{\circ} & 0 \\
0 & 0 & 1
\end{bmatrix}\cdot
\begin{bmatrix}
\Delta a \\
0 \\
0
\end{bmatrix}[/math]
[math]\begin{bmatrix}
x'' \\
y'' \\
z''
\end{bmatrix}=
\begin{bmatrix}
\Delta a\ cos\ 6^{\circ} \\
\Delta a\ sin\ 6^{\circ} \\
0
\end{bmatrix}[/math]
[math](x'', y'', z'')_{center} = (\Delta a\ cos\ 6^{\circ} , \Delta a\ sin\ 6^{\circ} , 0 )= (h'', k'', 0) [/math]
Performing an active rotation, we will rotate the equation for an ellipse in the frame of the DC to the frame of the wires . In the frame of the DC, the ellipse is centered on the x' axis, with the intersection points not having a uniform spacing in the ellipse parameter. In the frame of the wires, the ellipse is tilted [math]6^{\circ}[/math] counterclockwise from the x axis, with the intersection points having a uniform spacing in the ellipse parameter.
[math]\begin{bmatrix}
x'' \\
y'' \\
z''
\end{bmatrix}=
\begin{bmatrix}
cos\ 6^{\circ} & -sin\ 6^{\circ} & 0 \\
sin\ 6^{\circ} &cos\ 6^{\circ} & 0 \\
0 & 0 & 1
\end{bmatrix}\cdot
\begin{bmatrix}
x' \\
y' \\
z'
\end{bmatrix}[/math]
[math]\begin{bmatrix}
x'' \\
y'' \\
z''
\end{bmatrix}=
\begin{bmatrix}
x'\ cos\ 6^{\circ}+y'\ sin\ 6^{\circ} \\
-x'\ sin\ 6^{\circ}+y'\ cos\ 6^{\circ} \\
0
\end{bmatrix}[/math]
Substituting this into the equation for an ellipse in the frame of the wires,
[math]\frac{(x''+h'')^2}{a^2}+\frac{(y''+k'')^2}{b^2}=1[/math]
[math]\frac{x'\ cos\ 6^{\circ}+y'\ sin\ 6^{\circ}+\Delta a ^2}{a^2}+\frac{-x'\ sin\ 6^{\circ}+y'\ cos\ 6 ^{\circ}^2}{b^2}=1[/math]