General Form for a Parametric Parabola This question is an inverse of this other question on the  Parametric Form for a General Parabola. 

What is the general form, ie. $(Ax+Cy)^2+Dx+Ey+F=0$ , for the parabola given in parametric form as follows:
  $$\big(at^2+bt+c,\;\; pt^2+qt+r\big)$$

In other words, find $A,C,D,E,F$ in terms of $a,b,c,p,q,r$. 
I've posted a solution, but would like to see other approaches to this problem.

NOTE (Added April 2018)
See Latest Addendum to solution which provides Cartesian equation in neat matrix determinant form. 
 A: LATEST UPDATE (April 2018)
Using the method here, the Cartesian form of the parametric parabola 
$$x=at^2+bt+c\\
y=pt^2+qt+r $$
may be written in very neat and compact form using matrix notation as follows:
$$\color{red}{\left|\;\;\begin{matrix}
\left|\;\;\;\begin{matrix}
-a&&-p\\
-b&&-q
\end{matrix}\;\;\;\right|
&&\left|\begin{matrix}
-a&-p\\
(x-c)&(y-r)\end{matrix}\right|
\\\\
\left|\begin{matrix}
-a&-p\\
(x-c)&(y-r)\end{matrix}\right|
&&\left|\begin{matrix}
-b&-q\\
(x-c)&(y-r)\end{matrix}\right|
\end{matrix}\;\;\right|=0}$$
A simple derivation is as follows:
$$\begin{align}
x-c&=at^2+bt=t(at+b)\tag{1}\\
y-r&=pt^2+qt=t(pt+q)\tag{2}\\\\
p\cdot (1)-a\cdot (2):\qquad
p(x-c)-a(y-r)&=(pb-aq)t\tag{3}\\
q\cdot (1)-b\cdot (2):\qquad
q((x-c)-b(y-r)&=(aq-p)t^2\tag{4}\\\\
(3)^2/4:\qquad
\frac {[p(x-c)-a(y-r)]^2}{q(x-c)-b(y-r)}&=-(pb-aq)\\\\
[p(x-c)-a(y-r)]^2&=(aq-pb)[q(x-c)-b(y-r)]\\\\
\left|\begin{matrix}-a&-p\\(x-c)&(y-r)\end{matrix}\right|^2
&=\left|\begin{matrix}-a&-p\\-b&-q\end{matrix}\right|\;
\left|\begin{matrix}-b&-q\\(x-c)&(y-r)\end{matrix}\right|\\\\
\left|\;\;\begin{matrix}
\left|\;\;\;\begin{matrix}
-a&&-p\\
-b&&-q
\end{matrix}\;\;\;\right|
&&\left|\begin{matrix}
-a&-p\\
(x-c)&(y-r)\end{matrix}\right|
\\\\
\left|\begin{matrix}
-a&-p\\
(x-c)&(y-r)\end{matrix}\right|
&&\left|\begin{matrix}
-b&-q\\
(x-c)&(y-r)\end{matrix}\right|
\end{matrix}\;\;\right|&=0
\end{align}$$
See desmos implementation here.

ORIGINAL SOLUTION (Dec 2016)
Put $h=x-c,\;\; k=y-r$ and $2u=b,\;\; 2v=q$. 
Solving the quadratic for $t$ for both components and equating gives
$$\pm \ p\sqrt{u^2+ah}\;\;\mp a\sqrt{v^2+pk}=pu-av$$
Squaring and rearranging terms gives
$$
\small\begin{align}
(hp-ka)^2+4(pu-av)(hv-ku)&=0\\\\
(hp-ka)^2+(pb-aq)(hq-kb)&=0\\\\
(px-ay)^2\\ -2(pc-ar)(px-ay)+pc-ar)^2\\ +(pb-aq)[(qx-by)-qc-br)]&=0\\\\
\color{red}{(px-ay)^2}\\
\color{red}{\overbrace{+\big[q(pb-aq)-2p(pc-ar)\big]}^D \; x}\\
\color{red}{\overbrace{-\big[b(pb-aq)-2a(pc-ar)\big]}^E \; y}\\
\color{red}{\overbrace{+(pc-ar)^2-(pb-aq)(qc-br)}^F}&\color{red}{=0}
\end{align}$$
i.e. the parabola $$\big(at^2+bt+c,\; pt^2+qt+r\big)$$ can be written as 
$$(Ax+Cy)^2+Dx+Ey+F=0$$ where
$$\begin{align}
A&=\;\;\;p\\
C&=-\;a\\
D&=\;\;\;q(pb-aq)-2p(pc-ar)\\
E&=-\big[b(pb-aq)-2a(pc-ar)\big]\\
F&=\;\;(pc-ar)^2-(pb-aq)(qc-br)
\end{align}$$
See graphical implementation here.
Further using vector notation, we can write
$$\begin{align}
\mathbf{m}&=[p\;\;\; -a]\\
\mathbf{n}&=[b\qquad q]\\
\mathbf{k}&=[c\qquad r]\\
\mathbf{x}&=[x\qquad y]\end{align}$$
in which case we then have
$$\begin{align}
\mathbf{m\cdot x}&=px-ay\\
\mathbf{m\cdot n}&=pb-aq\\
\mathbf{m\cdot k}&=pc-ar\\
\end{align}$$
The general form can then be written as
$$\color{red}{(\mathbf{m\cdot x})^2
+\mathbf{\big[m\cdot} \big(q\ \mathbf{n}-2p\ \mathbf{k}\big)
\;\;\;
-\mathbf{m\cdot}\big(b\ \mathbf{n}-2a\ \mathbf{k}\big)\big]\mathbf{\cdot x}
+(\mathbf{m\cdot k})^2-(qc-br)\ \mathbf{m\cdot n}=0}
$$

Addendum (added Apr 2018)
The standard Cartesian can also be written more neatly using matrix determinant notation as follows:
$$\color{red}{(px-ay)^2
+
\overbrace{
\left|\begin{matrix}
q&2p \\
\left|\begin{matrix}p&r\\a&c\end{matrix}\right|
&\left|\begin{matrix}p&q\\a&b\end{matrix}\right|
\end{matrix}\right|
}^{D}\;x
-
\overbrace{
\left|\begin{matrix}
b&2a \\
\left|\begin{matrix}p&r\\a&c\end{matrix}\right|
&\left|\begin{matrix}p&q\\a&b\end{matrix}\right|
\end{matrix}\right|
}^{E}
\; y
-
\overbrace{
\left|\begin{matrix}
\left|\begin{matrix}q&r\\b&c\end{matrix}\right|
&\left|\begin{matrix}p&r\\a&c\end{matrix}\right|\\
\left|\begin{matrix}p&r\\a&c\end{matrix}\right|
&\left|\begin{matrix}p&q\\a&b\end{matrix}\right|
\end{matrix}\right|
}^F
=
0}
$$
The Cartesian form of the parabola can also be written in matrix form as follows:
$$\color{red}{\begin{matrix}
\big(x&y&1\big)\\\\\\\\\\\end{matrix}\;\;
\left(\begin{matrix}
p^2
 &-ap
  &\;\;\scriptsize\frac 12\left|\begin{matrix} Q &2R\\p&q\end{matrix}\right|\\
-ap
 &a^2
    &\scriptsize-\frac 12\left|\begin{matrix}Q &2R\\a&b\end{matrix}\right|\\
\scriptsize\frac 12\left|\begin{matrix}Q&2R\\p&q\end{matrix}\right|
 &\scriptsize-\frac 12 \left|\begin{matrix} Q &2R\\a&b\end{matrix}\right|
  &\;\;\;\scriptsize\left|\begin{matrix}R &\;Q\\S&\;R\end{matrix}\right|
\end{matrix}\right)\;\;
\left(\begin{matrix}x\\\\y\\\\1\end{matrix}\right)
=0}$$
where 
$$P=\left|\begin{matrix}x&y\\a&p\end{matrix}\right|=px-ay$$
$$Q=\left|\begin{matrix}b&q\\a&p\end{matrix}\right|=pb-aq$$
$$R=\left|\begin{matrix}c&r\\a&p\end{matrix}\right|=pc-ar$$
$$S=\left|\begin{matrix}c&r\\b&q\end{matrix}\right|=qc-br$$
Or better still,
$$\color{red}{\begin{matrix}
\big((x-c)&(y-r)&1\big)\\\\\\\\\\\end{matrix}\;\;
\left(\begin{matrix}
p^2
 &-ap
  &\;\;\scriptsize\frac q2\left|\begin{matrix} a &b\\p&q\end{matrix}\right|\\
-ap
 &a^2
    &\scriptsize-\frac b2\left|\begin{matrix}a&b\\p&q\end{matrix}\right|\\
\scriptsize\frac q2\left|\begin{matrix}a&b\\p&q\end{matrix}\right|
 &\scriptsize-\frac b2 \left|\begin{matrix} a&b\\p&q\end{matrix}\right|
  &\;\;\;0
\end{matrix}\right)\;\;
\left(\begin{matrix}x-c\\\\y-r\\\\1\end{matrix}\right)
=0}$$
