Intersection between two hyperboloids 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 $P_1, P_2, P_3$ :

How can I describe the intersection (in red below) between these two surfaces?
Maybe this can be done through parametrization?

 A: 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).
