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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?

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  • $\begingroup$ 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.) $\endgroup$ – David K Jan 21 '19 at 19:40
  • $\begingroup$ Indeed. (I guess I don't want the degenerate cases where the cone is flat or simply a line either.) $\endgroup$ – DominicR Jan 21 '19 at 19:58
  • $\begingroup$ 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. $\endgroup$ – amd Jan 21 '19 at 21:12
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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: $OP=d/\cos t$.

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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.

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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.

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