# When are cone geodesics planar

I mentioned to my (high school) students today that the intersection of a plane and a cone gives a conic section.

One asked whether if you 'unroll the cone' the conic section becomes a straight line on the resulting circular sector.

I can find examples that show this is false in general, but are there instances where it is true?

More specifically, given the cone defined by $$\alpha\rho=z$$ in cylindrical polars and the points $$A=(\rho_0,0,\alpha\rho_0), B=(\rho_0,\phi_0,\alpha\rho_0)$$, is the geodesic from $$A$$ to $$B$$ ever a planar curve?

• I suppose you are excluding degenerate conics where the plane of the section passes through the vertex of the cone in such a way that the conic section is a straight line in that plane. (This seems to be implied by your choice of notation.) – David K Jan 21 '19 at 19:40
• Indeed. (I guess I don't want the degenerate cases where the cone is flat or simply a line either.) – DominicR Jan 21 '19 at 19:58
• Try coming at it from the other direction: are there any line segments contained within a circular sector that are planar when “rolled up?” Any radius obviously qualifies, but those correspond to degenerate conics. – amd Jan 21 '19 at 21:12

Let line $$PH$$ be a geodesic line on the lateral surface of a cone (I'll suppose $$PH$$ does not pass through the vertex $$O$$ of the cone), and $$H$$ its point nearest to $$O$$. Let $$OH=d$$ and $$t=\angle POH$$. Then: $$OP=d/\cos t$$.

If $$a=OM=ON$$ is the slant height of the cone, then (see figures): $$\text{arc}\ MN = ta=\theta a\sin\alpha, \quad\text{and:}\quad \theta={t\over\sin\alpha},$$ where $$\alpha$$ is the semi-aperture of the cone. It follows that we can write the coordinates of point $$P$$ as: $$P(t)=\left({d\over\cos t}\sin\alpha\cos\left({t\over\sin\alpha}\right), {d\over\cos t}\sin\alpha\sin\left({t\over\sin\alpha}\right), {d\over\cos t}\cos\alpha\right).$$ The above equation defines then the curve in space corresponding to geodesic $$PH$$; parameter $$t$$ can take any value in $$(-\pi/2, \pi/2)$$.

This is a plane curve only if its torsion $$\tau(t)$$ vanishes for all $$t$$. But a straightforward calculation gives: $$\tau(t)={1\over d}{\cot\alpha \sin t \cos^2 t},$$ which vanishes only for $$t=0$$. Hence it cannot be a plane curve.

From differential geometry ( may not be suitable for high school students) consider the definition of geodesic torsion ( $$\psi$$ is angle between conic arc and slant generator). The curve is planar when torsion vanishes.

$$\tau_g = (\kappa_1-\kappa_2) \sin \psi \cos \psi$$ where $$\kappa's$$ are principal curvatures

Since $$\kappa_1=0$$ and for straight generator minor curvature of Euler's relation

$$\kappa_2 = \frac {\cos \alpha}{r}$$ and since by Clairaut's Law

$$r_{min}= r\cdot \sin \psi$$

$$\tau_g= \frac{\cos \alpha}{r_{min} }\, \sin^2\psi \cos \psi$$

can vanishes only for $$\psi=0,\,$$ i.e, when intersection plane contains the cone apex, all other terms are non-zero.