TITLE: What is the equation of a hyperbola for a plane intersecting a $95^\circ$ cone at right angles?
QUESTION [5 upvotes]: Plane intersecting a $95^\circ$ cone at right angles:

I'm trying to build a homemade tracking platform for my telescope.  I'm hoping that I can come up with an equation that I can use to cut a bearing for it on a milling machine.  I think it is a hyperbola generated by the intersection of a plane to a cone.  The plane intersects the cone $34.25$ inches from the tip and at a right angle.  The cone has an angle of $95^\circ$ (twice my latitude).

REPLY [2 votes]: An equation of a right cone with apex at the origin, axis along the $z$-axis and aperture angle $2\phi$ is $x^2+y^2=z^2\tan^2\phi$, which can be represented by the matrix $Q=\operatorname{diag}(1,1,-\tan^2\phi,0)$. Setting $\mathbf x = (x,y,z,1)^T$, the equation can then be written in the matrix form $\mathbf x^TQ\mathbf x=0$.  
We can similarly write the parameterization $\mathbf p_0+s\mathbf u+t\mathbf v$ of a plane as $$\mathbf X = \begin{bmatrix}\mathbf u&\mathbf v&\mathbf p_0\\0&0&1\end{bmatrix} \begin{bmatrix}s\\t\\1\end{bmatrix} = M\mathbf x.$$ This amounts to defining a coordinate system on the plane with $\mathbf p_0$ as the origin. In this coordinate system, the conic defined by the intersection of the cone and plane has the equation $$(M\mathbf x)^TQ(M\mathbf x) = \mathbf x^T(M^TQM)\mathbf x= 0,$$ i.e., this conic has the matrix $M^TQM$.  
A simple and convenient choice for $M$ for this problem is $$M=\begin{bmatrix}0 & -\cos\phi & -d\sin\alpha \\ 1 & 0 & 0 \\ 0 & \sin\phi & -d\cos\alpha \\ 0&0&1 \end{bmatrix}.$$ This makes the $y$-axis on the plane the transverse axis of the hyperbola and places the origin at the vertex of the lower branch, which is a distance $d$ down the side of the cone. We then have $$M^TQM = \begin{bmatrix}1&0&0 \\ 0&\cos2\phi \sec^2\phi & d\tan\phi \\ 0&d\tan\alpha&0\end{bmatrix}.$$ This corresponds to the equation $$x^2+y^2\cos2\phi\sec^2\phi+2dx\tan\phi=0.$$ Using any of the usual methods we can find that the center of this conic is at $\left(0,-\frac d2\tan2\phi\right)$. Translating the origin to this point produces the equation $$x'^2 + y'^2\cos2\phi\sec^2\phi=d^2\sec2\phi\sin^2\phi.$$ For $\phi\gt\pi/4$, this is a hyperbola (although since $\phi$ is close to $\pi/4$ it is very close to a parabola). From this equation, we can find that the transverse half-axis length is $-\frac d2\tan2\phi$ and conjugate half-axis length is ${d\over\sqrt{1-\cot^2\phi}}$. In your case, $\phi=95°/2=47.5°$ and $d=34.25$, so the half-axis lengths work out to be approximately $195.74$ and $85.54$, respectively.