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.

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.
