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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?

pseudoeuclidean
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2 Answers2

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$C_1:(x+\frac{\xi}{2})^2+y^2=(\frac{z-\frac{\zeta}{2}}{\alpha})^2\\ C_2:(x+\frac{\xi}{2})^2+y^2=(\frac{z+\frac{\zeta}{2}}{\alpha})^2$

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

$z = \alpha \sec \theta\\ x =-\frac {\zeta}{\xi\alpha} \sec\theta\\ y = \left(\sqrt {(\frac {\xi}{2})^2-(\frac {\zeta}{2\alpha})^2}\right)\tan \theta$

Doug M
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Say we have two cones who's axis are $2d$ apart and with the vertex z coordinate $2r$ apart (in your problem $2d=\xi$ and $2r=\zeta$) I'll leave $\alpha=1$, but the math should be very similar. I tried Doug's method, but the intersection curve came out wrong, I think the trig sub was off. This method is similar and uses a different trig sub.

$C1:(x+d)^2+y^2=(z+r)^2 $

$C2:(x-d)^2+y^2=(z-r)^2 $

expand, subtract and solve for $z$

$z=\frac{xd}{r}$

substitute that back into $C1$

we get

$$x^2+2xd+d^2+y^2=\frac{d^2x^2}{r^2}+2xd+r^2$$ $$x^2[1-\frac{d^2}{r^2}]+y^2=r^2-d^2$$ $$\frac{x^2}{r^2}[r^2-d^2]+y^2=r^2-d^2$$ $$\frac{x^2}{r^2}+\frac{y^2}{r^2-d^2}=1$$

we see that the determining factor whether or not $r$ is greater than $d$. lets assume d>r and our intersection makes a hyperbola. Then we get $$\frac{x^2}{r^2}-\frac{y^2}{d^2-r^2}=1$$

We can write parametric equations for this curve with $sinh/cosh$ or $sec/tan$ if we do the latter. $$x=r\sec(\theta)$$ $$y=\sqrt{(d^2-r^2)}\tan(\theta)$$ $$z=d\sec(\theta)$$

two cones with red intersection curve

k2shah
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