I have two one-sheet hyperboloids defined as locus of points with same difference of distances from two arbitrary points in space (the foci).

Given $P_1 (x_1, y_1, z_1)$, $P_2 (x_2, y_2, z_2)$ and the euclidean distance between two points $d=\sqrt{(x_a-x_b)^2+(y_a-y_b)^2+(z_a-z_b)^2}$

I can define the hyperboloid with difference of distances $d_{1,2} \in \mathbb{R}$ as

$\sqrt{(x-x_1)^2+(y-y_1)^2+(z-z_1)^2} - \sqrt{(x-x_2)^2+(y-y_2)^2+(z-z_2)^2} = d_{1,2}$

Same for another hyperboloid with difference of distances $d_{1,3} \in \mathbb{R}$ from $P_1$ and $P_3(x_3, y_3, z_3)$

$\sqrt{(x-x_1)^2+(y-y_1)^2+(z-z_1)^2} - \sqrt{(x-x_3)^2+(y-y_3)^2+(z-z_3)^2} = d_{1,3}$

(By the way, is this correct or there are neater ways to describe this kind of hyperboloids?)

This is a plot for some negative values $d_{1,2}\; d_{1,3}$ and some coordinates for $P_1, P_2, P_3$ :

enter image description here

How can I describe the intersection (in red below) between these two surfaces?

Maybe this can be done through parametrization?

enter image description here

  • $\begingroup$ From your diagram $P_1$ seems to be in the interior of both hyperboloids, but if $a$ and $b$ are positive the equations you wrote would imply the opposite. $\endgroup$ – Aretino Jan 31 '18 at 17:33
  • $\begingroup$ @Aretino thank you for your comment, I'll update the domain for $a$ and $b$, they have no restrictions. In the example graph I've chosen two negative values for them. Maybe the fact that I called the difference of distances $2a$ is misleading, I can call it just $d_1$ $\endgroup$ – Guglie Jan 31 '18 at 17:43
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    $\begingroup$ Writing the equations of the surfaces as two quadrics, and subtracting one of them from the other, will give you the equations of two planes. The intersection between the hyperboloids is then the same as the intersection between one of the hyperboloids and one of the planes, i.e. a conic. $\endgroup$ – Aretino Jan 31 '18 at 18:24
  • $\begingroup$ Thanks again @Aretino, how can I write those surfaces as quadrics? I've asked another question on this subproblem, maybe you can reply there math.stackexchange.com/q/2630471/526162 $\endgroup$ – Guglie Jan 31 '18 at 23:23
  • $\begingroup$ I could go into more details if I were sure that this is what are looking for. Your question "How can I describe the intersection between these two surfaces?" is too vague: a conic section can be described as the intersection between a plane and a quadric (but I don't know if that is enough for you) or as a parametric curve. What do you need exactly? $\endgroup$ – Aretino Feb 1 '18 at 15:11

Rewrite your first hyperboloid equation as $$ \sqrt{(x-x_1)^2+(y-y_1)^2+(z-z_1)^2} - d_{12}= \sqrt{(x-x_2)^2+(y-y_2)^2+(z-z_2)^2}. $$ Square both sides and rearrange to get $$ \tag{1} 2x(x_2-x_1)+2y(y_2-y_1)+2z(z_2-z_1) +d_1^2 - d_2^2 + d_{12}^2= 2d_{12}\sqrt{(x-x_1)^2+(y-y_1)^2+(z-z_1)^2}, $$ where $d_1^2=x_1^2+y_1^2+z_1^2$ and $d_2^2=x_2^2+y_2^2+z_2^2$. Do the same for the other hyperboloid: $$ \tag{2} 2x(x_3-x_1)+2y(y_3-y_1)+2z(z_3-z_1) +d_1^2 - d_3^2 + d_{13}^2= 2d_{13}\sqrt{(x-x_1)^2+(y-y_1)^2+(z-z_1)^2}, $$ where $d_3^2=x_3^2+y_3^2+z_3^2$. If you now multiply $(1)$ by $d_{13}$, $(2)$ by $d_{12}$, and subtract $(2)$ from $(1)$, you end up with a linear equation: $$ \tag{3} 2x[d_{13}(x_2-x_1)-d_{12}(x_3-x_1)]+ 2y[d_{13}(y_2-y_1)-d_{12}(y_3-y_1)]+ 2z[d_{13}(z_2-z_1)-d_{12}(z_3-z_1)] +d_{13}(d_1^2 - d_2^2 + d_{12}^2) -d_{12}(d_1^2 - d_3^2 + d_{13}^2)= 0. $$ This is the equation of a plane: the locus you need is then the intersection between this plane and either hyperboloid, that is a conic section (hyperbola or ellipse)

As you can see the equations are getting more and more complicated, so I will just sketch a possible strategy to parameterize that conic.

  • Square equation $(1)$ (or $(2)$, it doesn't matter) and simplify the result.

  • Insert into that equation the expression for $z$ obtained from $(3)$ (if equation $(3)$ doesn't contain $z$ just solve for $x$ or $y$ instead).

  • The equation you thus get is that of a conic section in plane $xy$, projection onto that plane of the original conic. You can find its center and axes with the usual formulas and thus find parametric equations $x(t)$ and $y(t)$ for its coordinates.

  • Plug $x(t)$ and $y(t)$ into the expression for $z$ obtained above, to get also z(t).


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