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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.

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  • $\begingroup$ The answers to this question describe several ways to do this. There’s also an old question about how to convert a quadratic Bézier curve into a Cartesian equation, to which this problem can be converted by setting the control points to $P_0(c,r)$, $P_1(b/2+c,q/2+r)$ and $P_2(a+b+c,p+q+r)$. $\endgroup$ – amd Dec 12 '16 at 7:39
  • $\begingroup$ @amd - That's an interesting reference. $\endgroup$ – hypergeometric Dec 12 '16 at 7:49
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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,

$$\color{red}{\begin{matrix} \big((x-c)&(y-r)&1\big)\\\\\\\\\\\end{matrix}\;\; \left(\begin{matrix} p^2 &-ap &\;\;\scriptsize\frac q2\left|\begin{matrix} a &b\\p&q\end{matrix}\right|\\ -ap &a^2 &\scriptsize-\frac b2\left|\begin{matrix}a&b\\p&q\end{matrix}\right|\\ \scriptsize\frac q2\left|\begin{matrix}a&b\\p&q\end{matrix}\right| &\scriptsize-\frac b2 \left|\begin{matrix} a&b\\p&q\end{matrix}\right| &\;\;\;0 \end{matrix}\right)\;\; \left(\begin{matrix}x-c\\\\y-r\\\\1\end{matrix}\right) =0}$$

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  • $\begingroup$ $\mathbf m$ is the normal to the parabola’s axis, but what geometric meaning do the others have, if any? It would be nice if they had one instead of only being a notational convenience. $\endgroup$ – amd Dec 12 '16 at 17:49
  • $\begingroup$ @amd - Please share your views if you know the answer. $\endgroup$ – hypergeometric Dec 13 '16 at 5:01

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