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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.


marked as duplicate by amd, Namaste, user91500, Daniel W. Farlow, Vladimir Vargas Dec 12 '16 at 23:22

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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

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,

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

  • $\begingroup$ $\mathbf m$ 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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