<s>[INST] Parameterization of the Intersection of Two Parallel Cones I need to parameterize the curve of intersection of two congruent, parallel cones, where $\xi$ is the distance between the axes of symmetry, $\zeta$ is the difference in $z$ position of the vertices, and $\alpha$ is the slope of the cones.
I wrote the equations for the two cones below. $\xi$ and $\zeta$ are both divided by two to keep the problem symmetric about the origin.
$$C_1:(x-\frac{\xi}{2})^2+y^2=(\frac{z-\frac{\zeta}{2}}{\alpha})^2$$
$$C_2:(x+\frac{\xi}{2})^2+y^2=(\frac{z+\frac{\zeta}{2}}{\alpha})^2$$
Normally, to find the intersection of any two equations, I would solve the two equations for the same variable or constant and set the equations equal to eachother. In the case of this problem, it is easy to solve both equations for $y^2$, yielding:
$$C_1:y^2=(x-\frac{\xi}{2})^2-(\frac{z-\frac{\zeta}{2}}{\alpha})^2$$
$$C_2:y^2=(x+\frac{\xi}{2})^2-(\frac{z+\frac{\zeta}{2}}{\alpha})^2$$
Then
$$(x-\frac{\xi}{2})^2-(\frac{z-\frac{\zeta}{2}}{\alpha})^2=(x+\frac{\xi}{2})^2-(\frac{z+\frac{\zeta}{2}}{\alpha})^2$$
I am hesitant to continue, because I just eliminated $y$ from the equation. Intuitively, I know that $y$ should persist throughout the problem, because the curve will occupy the $x$, $y$, and $z$ space.
Am I going about this problem incorrectly? Can someone steer me in the right direction?
 [/INST] $C_1:(x+\frac{\xi}{2})^2+y^2=(\frac{z-\frac{\zeta}{2}}{\alpha})^2\\
C_2:(x+\frac{\xi}{2})^2+y^2=(\frac{z+\frac{\zeta}{2}}{\alpha})^2$
Note, if $|\frac {\zeta}{\alpha}| > \xi$ then we have the cones nested inside each other.  Since they are double cones there will still be a curve of intersection.  If they are nested we have an ellipse.
If they are not. We get something hyperbolic.
If the solution is the ellipse we want $x,y,z$ in terms of $\sin\theta,\cos\theta$
If the solution is the hyperbola we want $x,y,z$ in terms of $\sinh\theta,\cosh\theta$ or $\tan\theta, \sec\theta$
Multiply them out:
$x^2 + x\xi + \frac{\xi^2}{4}+y^2=\frac{z^2-z\zeta+\frac {\zeta^2}{4}}{\alpha^2}\\
x^2 - x\xi + \frac{\xi^2}{4}+y^2=\frac{z^2+z\zeta+\frac {\zeta^2}{4}}{\alpha^2}$
Subtract one from the other
$2x\xi = -\frac {2z\zeta}{\alpha^2}\\
x = -\frac {z\zeta}{\xi\alpha^2}$
Substitute into either of the original surfaces to find $y$ in terms of  $z.$
$y^2 = $$(\frac {z + \frac {\zeta}{2}}{\alpha})^2 - (\frac {-z\zeta}{\xi\alpha^2}-\frac{\xi}{2})^2\\
\frac {z^2 +z\zeta + \frac {\zeta^2}{4}}{\alpha^2} - (\frac {z^2\zeta^2}{\xi^2\alpha^4} +\frac {z\zeta}{\alpha^2}+\frac {\xi^2}{4})$
$4\xi^2\alpha^4 y^2 = $$(\xi^2\alpha^2-\zeta^2) z^2+\alpha^2(\zeta^2 - \alpha^2\xi^2)\\
(\xi^2\alpha^2-\zeta^2) (z^2-\alpha^2)$
$y^2 = {((\frac {\xi}{2})^2- (\frac {\zeta}{2\alpha})^2)((\frac {z}{\alpha})^2-1)}$
if $(\frac {\xi}{2})^2- (\frac {\zeta}{2\alpha})^2 < 0$ we have the elipse.
$z = \alpha\cos \theta\\
x =-\frac {\zeta}{\xi\alpha} \cos\theta\\
y = \left(\sqrt {(\frac {\zeta}{2\alpha})^2-(\frac {\xi}{2})^2}\right)\sin \theta$
if $(\frac {\xi}{2})^2- (\frac {\zeta}{2\alpha})^2 > 0$ we have the hyperbola
$z = \alpha \sec \theta\\
x =-\frac {\zeta}{\xi\alpha} \sec\theta\\
y = \left(\sqrt {(\frac {\xi}{2})^2-(\frac {\zeta}{2\alpha})^2}\right)\tan \theta$
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