Is the following a conic section All vectors are in $\mathbb{R}^3$ and only $\mathbf{r} = \left[ x; y; z \right]$ is unknown. My question is does the following system define a conic section in the $x-y$ plane and, if so, how can I find it:
$$
\begin{align}
\mathbf{r}^\mathrm{T} \left[ \begin{array}{c} 0 \\ 0 \\ 1 \end{array} \right]
 & = 0 \\
\\
\mathbf{v}_1^\mathrm{T} \frac {\mathbf{r} - \mathbf{r}_1} {\Vert \mathbf{r} - \mathbf{r}_1 \Vert } + \mathbf{v}_2^\mathrm{T} \frac {\mathbf{r} - \mathbf{r}_2} {\Vert \mathbf{r} - \mathbf{r}_2 \Vert } & = c
\end{align}
$$
If either $\Vert \mathbf{v}_1 \Vert = 0$ or $\Vert \mathbf{v}_2 \Vert = 0$, then the above is the intersection of a cone and the $z=0$ plane. Likewise if $\mathbf{r} = \mathbf{r}_1$ or $\mathbf{r} = \mathbf{r}_2$. However, I have been unable to figure out what the above represents in the general case.
 A: No, these are not conic sections in general.  
Your first equation forces $z = 0$.  For \begin{align*}
\mathbf{r}_1 = \mathbf{v}_2 = \begin{pmatrix} 1 \\ 0 \\ 0 \end{pmatrix} \\ 
\mathbf{r}_2 = \mathbf{v}_1 = \begin{pmatrix} 0 \\ 1 \\ 0 \end{pmatrix}  \text{,}
\end{align*}
with $c = 0$, your second equation reduces to 
$$  \frac{x}{\sqrt{|x|^2+|y-1|^2}} + \frac{y}{\sqrt{|x-1|^2+|y|^2}} = 0  \text{.}  $$
This is not the equation of a conic section.  It is the line $y = 1-x$ for $x \in (-\infty, 0] \cup [1,\infty)$ joined by the arc of the circle centered at $(1/2,1/2)$ with radius $1/2$ for the angles in the interval $[3\pi/4, 7\pi/4]$.

Changing only $c=3/2$, it's not a conic section.

Changing $\mathbf{v}_1 = \begin{pmatrix}0\\2\\0\end{pmatrix}$ and setting $c = 1/2$, the graph loses its symmetry.

The $r_i$ seems to control where the pieces of the piecewise curve meet.  Many choices of $c$ give no solutions (which is another thing that cannot happen with a planar section of nappes).
Why would you believe there is a short description of what is represented by varying the parameters in your equation?
A: If $r_1=r_2$ then it is clear that it is a comic section because 
$(v_1^T+v_2^T)(r-r_1)=c||r-r_1||$
so
$c^2||r-r_1||^2-((v_1^T+v_2^T)(r-r_1))^2=0$
that is an equation of a polynomial of order $2$
In general you have 
$||r-r_2||v_1^T(r-r_1)+||r-r_1||v_2^T(r-r_2)=c||r-r_1||||r-r_2||$
so
$(||r-r_2||v_1^T(r-r_1)+||r-r_1||v_2^T(r-r_2))^2-(c||r-r_1||||r-r_2||)^2=0$
The problems are the factors 
$(||r-r_2||v_1^T(r-r_1))^2$
$(||r-r_1||v_2^T(r-r_2))^2$
and $(||r-r_2||v_1^T(r-r_1)) (||r-r_2||v_1^T(r-r_1))$
The first two factors are polynomials of order $3$ while in the last factor there is a root square.
