Cone Section: Intersection between a plane ax + by + cz = d and two cones of height $h$, raidus $r$ whose apexes lie at the centre, along the $z$ axis A double cone can be described by circles that increase with $|z|$ and have radius $R$ at $h = |z_R|$. It holds:
\begin{align}
    \frac{R^2}{h^2}\,z^2 = x^2 + y^2
\end{align}
The plane
\begin{align}
    ax + by + cz = d
\end{align}
should now intersect this cone. To do this, we move the plane to $z$ and insert this into the double cone equation.
\begin{align}
    &z = \frac{d - ax - by}{c} \\
&\Longrightarrow \hspace{10pt} \underbrace{\frac{R^2}{c^2h^2}}_{\color{red}{=\,\kappa^2}}\,(d-ax-by)^2 = x^2 + y^2 \\
&\Longrightarrow \hspace{10pt} \kappa^2\,(d^2+a^2x^2+b^2y^2-2adx-2bdy + 2abxy) = x^2 + y^2 \\
&\Longrightarrow \hspace{10pt} \left(\kappa^2a^2-1\right)x^2+\left(\kappa^2b^2-1\right)y^2 + \left(2\kappa^2 ab\right)xy + \left(-2\kappa^2ad\right)x + \left(-2bd\right)y + \kappa^2d^2 = 0
\end{align}
Unfortunately, this is not yet the right solution. If I take, for example, the plane
\begin{align}
    -h \cdot x + 0 \cdot y + 2R \cdot z = Rh
\end{align}
then I get:
\begin{align} 
     &\kappa^2 = \frac{1}{4h^2} \\
     &\Longrightarrow \hspace{10pt} \left(\frac{1}{4h^2}h^2-1\right)x^2-y^2 + \left(-2\,\frac{1}{4h^2}\,(-h)\,Rh\right)x + \frac{1}{4h^2} R^2h^2 = 0 \\
     &\Longrightarrow \hspace{10pt} -\frac{3}{4}x^2-y^2 + \frac{R}{2}\,x + \frac{R^2}{4} = 0 \\
     &\Longrightarrow \hspace{10pt} \frac{3}{R^2}x^2 + \frac{4}{R^2}y^2 - \frac{2}{R}\,x = 1 \\
\end{align}
And I also know where the problem is: I am supposed to look at the function in the sectional plane. So somehow I have to transform my coordinates from the original coordinate system into a coordinate system of the plane. Unfortunately, I don't know how. Maybe the task would even be easier if I had transformed my system right at the beginning. This way I would only have to consider my cone at $z=0$. My approach so far has been as follows:
My origin should be the point $(0,0,\frac{d}{c})$. My new $z$-axis is defined by the normal of the plane, so $\vec{z}_{new} = \frac{\vec{n}}{|\vec{n}|}$ with $\vec{n} = (a,b,c)$. My new $x$-axis shall continue to be perpendicular to the $y$-axis, so $\vec{x}_{new} = (\xi_1,0,\xi_2)$. My new $y$-axis is to be formed by the cross product of these two. Unfortunately, that's as far as I've got so far.

First, you need to express the unit normal to the plane in spherical coordinates:
$n = \dfrac{1}{\sqrt{a^2 + b^2+c^2}} (a, b, c) = (\sin \theta \cos \phi, \sin \theta \sin \phi, \cos \theta) $
Comparing the expressions, we have
$ \theta = \cos^{-1} \dfrac{c}{\sqrt{a^2 + b^2+c^2} }$
$ \phi = \text{ATAN2}(a,b) $
It follows that the steepest descent direction that lies in the plane is
$ u_1 = (\cos \theta \cos \phi, \cos \theta \sin \phi, - \sin \theta ) $
We can take $u_1$ as the $x$ axis in the plane, while the $y$ axis is given by
$ u_2 = (-\sin \phi, \cos \phi , 0 ) $
Using the $u_1, u_2$ axes, any point on the plane can be expressed as
$ r = (x, y, z) = r_0 + u_1 x_1 + u_2 x_2 $
Where $r_0$ is any point on the plane, for example $(0, 0, d / c )$
Written compactly,
$r = r_0 + V u \hspace{48pt} (1)$
with $V = [u_1, u_2] $ and $ u = [x_1, x_2]^T $
Now the equation of the cone is (in terms of $r$)
$ r^T Q r = 0 \hspace{48pt} (2)$
where
$Q = \begin{bmatrix} 1 && 0 && 0 \\ 0 && 1 && 0 \\ 0 && 0 && - \tan^2 \theta_c \end{bmatrix} $
Where $\theta_c$ is the semi-vertical angle of the cone, i.e. $\tan \theta_c = \dfrac{R}{h} $
Substitute $(1)$ into $(2)$
$ (r_0 + V u)^T Q (r_0 + V u) = 0 $
Expand
$ r_0^T Q r_0 + 2 u^T V^T Q r_0 + u^T V^T Q V u = 0 $
Let's find the matrix V^T Q ,
$V^T Q = \begin{bmatrix} \cos \theta \cos \phi && \cos \theta \sin \phi && \tan^2 \theta_c \sin \theta \\ - \sin \phi && \cos \phi && 0 \end{bmatrix} $
Therefore,
$V^T Q V = \begin{bmatrix} \cos^2 \theta - \tan^2 \theta_c \sin^2 \theta && 0 \\ 0 && 1 \end{bmatrix}$
If $ \cos^2 \theta - \tan^2 \theta_c \sin^2 \theta = 0 $, then we have a parabola.
If $ \cos^2 \theta - \tan^2 \theta_c \sin^2 \theta \gt 0 $ then $V^T Q V$ is invertible, and the conic section is an ellipse.
If $ \cos^2 \theta - \tan^2 \theta_c \sin^2 \theta \lt 0 $ then $V^T Q V$ is invertible, and the conic section is an hyperbola.
Let's do the second case.  We can find the center of this ellipse in $u_1u_2$ plane as follows:
$u_0 = - (V^T Q V)^{-1} V^T Q r_0 $
Explicitly evaluating $u_0$, we find
$u_0 = \begin{bmatrix} \dfrac{(-d/c) (\tan^2 \theta_c \sin \theta)}{\cos^2 \theta - \tan^2 \theta_c \sin^2 \theta} \\ 0 \end{bmatrix} $
Then by completing the square, we can write,
$(u - u_0)^T V^T Q V (u - u0) = u_0^T V^T Q V u_0 - r_0^T Q r_0 $
Now, $u_0^T V^T Q V u_0 = \dfrac{(d/c)^2 \tan^4 \theta_c \sin^2 \theta}{\cos^2 \theta - \tan^2 \theta_c \sin^2 \theta}$
and $r_0^T Q r_0 = -(d/c)^2 \tan^2 \theta_c $
Hence,
$u_0^T V^T Q V u_0 - r_0^T Q r_0 = \dfrac{(d/c)^2}{\cos^2 \theta - \tan^2 \theta_c \sin^2 \theta} ( \tan^4 \theta_c \sin^2 \theta + tan^2 \theta_c \cos^2 \theta - \tan^4 \theta_c \sin^2 \theta) $
And this simplifies to,
$ \text{Right Hand Side} = \dfrac{(d/c)^2}{\cos^2 \theta - \tan^2 \theta_c \sin^2 \theta} \cos^2 \theta \tan^2 \theta_c $
Therefore, the semi-major axis of the ellipse is along $u_1$ and given by,
$ a = (d/c) \dfrac{\cos \theta \tan \theta_c }{\cos^2 \theta - \tan^2 \theta_c \sin^2 \theta }$
And the semi-minor axis of the ellipse is along $u_2$ and given by,
$ b = (d/c) \dfrac{\cos \theta \tan \theta_c }{\sqrt{\cos^2 \theta - \tan^2 \theta_c \sin^2 \theta } } $
