Angles $(\psi -\theta) $ relation is unique for Conics in a 3D situation.
// It is to seek $r- \psi$ of planar sections aka Conics on Gauss curvature $K=0$ cones for uniqueness and generality. Motivation/Background is mentioned although not absolutely necessary for the question at present. It needed mentioning due to the simplicity of their relationship in deeper contexts.
Angle $\psi $ is reckoned counterclockwise direction consistently from radius vector to one of parallel geodesics. Constant rotation $\phi$ and curvature in two dimensions shown here:
Elliptic geometry
$$ \phi_{Elliptic}= \psi +\theta = const $$
Differentiating with respect to arc for elliptic straight lines
$$ \psi^{'}+\frac{d\theta}{ds} = \psi^{'}+\frac{\sin \psi}{r} =\kappa_{gElliptic}=0$$
Re-integrate we get Clairaut's constant in polar coordinates:
$$ r\cdot \sin \psi = r_{min} = OE $$
By Inversion
the angle $\psi$ remains constant in magnitude but changes sense/relative direction into
Hyperbolic geometry
$$ \phi_{Hyperbolic}= (\pi-\psi ) +\theta = const $$
Differentiating with respect to arc for hyperbolic straight lines
$$ \psi^{'}-\frac{d\theta}{ds} =\psi^{'}-\frac{\sin \psi}{r} =\kappa_{gHyperbolic} =0$$
Re-integrate we get constant ;
$$ r/\sin \psi = r_{max} = OH $$
$ (r, \theta, z) $ are cylindrical coordinates and tangent at $P$ to conic makes angle $\psi $ to cone generator $OP$ to compare in this form their projections with elliptic and hyperbolic geodesics in two dimensions. //
If it is constant for example when $ \psi = \pi/2 \text{ at } G $ then we have a circle formed by cutting a cone of any semi-vertical angle with a plane perpendicular to the cone axis.
How to establish/ find this relation?
Thanks for help in finding a reference or hints.
Best Answer
It's not really clear what kind of relation OP expects, but here's something ...
Let the generators of a cone (with vertex at the origin and $z$-axis as axis) make angle $\alpha$ with the $xy$-plane. Let a cutting plane parallel to the $y$-axis make angle $\beta$ with the $xy$-plane, and let it meet the cone at $V := (-v \cos\alpha,0,v \sin\alpha)$ (which will be a vertex of the conic). That is, the cone and plane have equations $$z^2 \cos^2\alpha = (x^2+y^2)\sin^2\alpha \qquad\qquad (x+v\cos\alpha)\sin\beta=(z-v\sin\alpha)\cos\beta$$ We can parameterize the intersection of the plane and cone by $$P := p (\cos\alpha \cos\theta, \cos\alpha\sin\theta,\sin\alpha)$$ with $\theta$-dependent $p$ determined by substituting $P$ into the plane equation. The fully-parametric form of $P$ is then
$$P = \frac{v \sin(\alpha+\beta)}{\sin\alpha\cos\beta - \cos\alpha\sin\beta \cos\theta}\,(\cos\alpha \cos\theta, \cos\alpha\sin\theta,\sin\alpha)$$ A tangent vector $P'$ is proportional to the derivative of $P$ with respect to $\theta$; in particular, we can take $$P' = ( \sin\alpha\cos\beta\sin\theta, \cos\alpha\sin\beta-\sin\alpha\cos\beta\cos\theta, \sin\alpha \sin\beta \sin\theta)$$ Then, the angle $\psi$ between the generator $OP$ and tangent vector $P'$ satisfies $$\cos\psi = \frac{P\cdot P'}{|P|\,|P'|}$$ We have $$\begin{align} P\phantom{^\prime}\cdot P' &= \frac{v \sin\beta \sin(\alpha+\beta) \sin\theta}{ \sin\alpha\cos\beta - \cos\alpha\sin\beta\cos\theta} \\[4pt] |P\phantom{^\prime}|^2 = P\phantom{^\prime}\cdot P\phantom{^\prime} &= \frac{v^2 \sin^2(\alpha+\beta)}{(\sin\alpha\cos\beta - \cos\alpha\sin\beta\cos\theta)^2} \\[4pt] |P'|^2 = P'\cdot P' &= (\sin\alpha\cos\beta - \cos\alpha\sin\beta\cos\theta)^2 + \sin^2\beta \sin^2\theta \end{align}$$ Thus,
for appropriate choices of sign. $\square$
(I was hoping for something a little more elegant.)
Incidentally, one can show that, in general, the eccentricity of the conic is given by $e = \sin\beta/\sin\alpha$, but this observation doesn't seem to make the formulas appreciably better.
For the "standard" cone with $\alpha=\pi/4$, we have $$\cot\psi = \pm\frac{\sqrt{2}\,\sin\theta}{\cot\beta-\cos\theta}$$
For any cone cut by a horizontal plane, $\beta=0$ (giving a circle), we have $$\cot\psi = 0$$ so that $\psi$ is constantly $\pi/2$, which is geometrically obvious (as noted by OP).
For the standard cone cut by a vertical plane, $\beta=\pi/2$ (giving a rectangular hyperbola), we have $$\cot\psi = \pm \sqrt{2}\tan\theta \quad\to\quad \cot\theta\cot\psi = \pm\sqrt{2}$$ The reader can verify that this agrees with @Aretino's formula, with the condition $a=b$.
For the standard cone cut by a comparably-inclined plane, $\beta=\alpha=\pi/4$ (giving a parabola), we have $$\cot\psi = \pm\frac{\sqrt{2}\,\sin\theta}{1-\cos\theta} = \pm\sqrt{2}\,\cot\frac12\theta$$