# Curvature of projected conic sections

who can prove an easy and beautiful observation on a sheet of paper in a few lines? I have used a computer algebra system to verify (the possible input is contained in the "proof") the following.

Let $C\subset\mathbb R^3$ be a cone (quadric) with half-angle $0<\varphi<\tfrac{\pi}{2}$, $P\in C$ a point in distance $\overline{oP}>0$ to the apex $o$ of $C$ and let $E$ be a plane containing $P$ but not the line $oP$. Then the orthogonal projection of the conic section $C\cap E$ onto a plane $F$ orthogonal to $oP$ has curvature $\cot(\varphi)/\overline{oP}$ at the intersection of $F$ and $oP$.

Proof: Introducing an orthogonal coordinate system $(x,y,z)$ and appying an isometry, $C$ is the locus of $$x^2 + (y\cos(\varphi) + z\sin(\varphi))^2 = \tan(\varphi)^2(z\cos(\varphi) - y\sin(\varphi))^2,$$ with the apex $o$ equal to the origin and $P:=(0,0,-d)$ lying on the $z$-axis for some $d>0$. The plane $E$ is the locus of $$x\eta+y\xi-(z+d)=0$$ for some $\eta,\xi\in\mathbb R$. The conic section $C\cap E$ is locally around $P$ parametrized by the $x$-coordinate and we have $\tfrac{\partial y}{\partial x}(0)=0$ and $\tfrac{\partial^2 y}{\partial x^2}(0)=\cot(\varphi)/d$, using a computer algebra system. $\square$

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One curvature is zero and another tangential $\cot(\varphi)/\overline{oP}$.If an inclined plane makes $\theta$ to normal plane, then $R_{inclined} = R_{normal}\cdot\cos \theta$ and by drawing radius perpendicular to axis $r = \overline {oP} \sin \phi; \theta =\phi.$