I need to find the locus of points in the real $(x, y)$ plane, in parametric form, satisfied by the equation
\begin{equation}\det\begin{pmatrix} 0 & 1 & 1 & 1 & 1\\ 1 & 0 & c_1 & y & c_2\\ 1 & c_1 & 0 & c_3 & x \\ 1 & y & c_3 & 0 & c_4 \\ 1 & c_2 & x & c_4 & 0\\ \end{pmatrix} = 0\,, \end{equation} for $c_1$, $c_2$, $c_3$ and $c_4$ real and positive semidefinite constants.
This equation is quadratic in $x$ (and also in $y$). The locus of points usually comes in three pieces. But if the largest of the constant $c_4$ satisfies $\sqrt{c_4} > \sqrt{c_1} + \sqrt{c_2} + \sqrt{c_1}$ there is a fourth piece.
Question: How do I obtain the four parametric equations $(x(t), y(t))$ with each one describing a single piece for a given choice of $c_i$'s.
The picture below illustrates a case with four pieces (I obtained this by solving the above quadratic equation for y in terms of x and plotted the two solutions).
I tried all sorts of transformations, but couldn't get a nice clean form for the curves. Because of the clean and symmetric form of the equation, I really believe there is a solution to this problem. Would anyone know?
When I simplified the determinant using MAPLE, I got something like
$$\eqalign{ & -{1 \over 2}\det A = {x^2}y + x{y^2} - \left( {{c_1} + {c_2} + {c_3} + {c_4}} \right)xy \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, + \left( {{c_2} - {c_4}} \right)\left( {{c_1} - {c_3}} \right)x + \left( {{c_3} - {c_4}} \right)\left( {{c_1} - {c_2}} \right)y \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, + \left( {{c_1} - {c_2} - {c_3} + {c_4}} \right)\left( {{c_1}{c_4} - {c_3}{c_2}} \right) \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = P(x,y) \cr} \tag{1}$$
Observations
$1.$ A nice re-statement of your problem is to find parametric equations for the family of curves
$$Q(x,y)=x^2y+xy^2+axy+bx+cy+d=0\tag{2}$$
where $a$, $b$, $c$, and $d$ are arbitrary real constants.
$2.$ I think that form $(1)$ is the best that you can have since any parameterization may need different cases to consider. You can just see the equation
$$P(x,y)=0\tag{3}$$
as the implicit equation of your curves and it contains all of them in just one equation!
$3.$ You can solve for $y$ from $(2)$ to obtain some explicit equations in the form $y=f(x)$. However, you should discuss the sign of the discriminant $\Delta$ of the quadratic equation for the existence of real roots which will be frustrating since $\Delta=\Delta(c_1,c_2,c_3,c_4,x)$!
$4.$ Another important point is that there are in fact just two explicit formulas $y_i=f_i(x), i=1,2$ for the curves not three or four! One for the $\color{blue}{\text{Blue}}$ part and one for the $\color{red}{\text{Red}}$ part. Those parts that you see nothing are the parts that for the given values of $c_i$'s and $x$, there is no real root for the quadratic equation. To be more specific, in those parts
$$\Delta(c_1,c_2,c_3,c_4,x) \lt 0 \tag{4}$$
$5.$ There are some cases that we have real roots for all values of $x$. The following picture shows the case $c_1=c_4=0$ and $c_2=-c_3=-5$. In this case, we have
$$\begin{array}{} P(x,y)=x^2y+xy^2+25x+25y=0 \\ xy(y+x)+25(y+x)=0 \\ (y+x)(xy+25)=0 \\ y=-x \\ y=\frac{-25}{x} \end{array}$$