Macro to draw a parabola with pgf/TikZ

Question

I have been trying to build a macro \parabola{...} to draw a parabola passing through 3 coordinates with TikZ but without success.

For example

\parabola{A}{B}{C}

would draw the parabola interpolating the (x,y) coordinates (A), (B) and (C). I would like also to specify the style of the curve, the plot domain, etc.

The main problem I found is that I cannot figure out how to obtain the value of a given coordinate in dimensionless form (given a certain unit).

Answer 1

Here is one solution. It essentially solves the linear equation obtained from quadratic interpolation. It is important to note however that it is far from perfect. In particular there are strong constraints on possible points because of tikz computations limitations: the numbers in the computations must be small enough. This is clearly not what tikz is made for. Using other ways to obtain the coefficients would be better (sagetex or asymptote or others).

At least it is a nice example of the use of letin a tikz path.

I hope the computations are clear enough. The coefficients A, B and C are the coefficients of the quadratic polynomial Ax^2 + Bx + C.

The points must be entered in increasing order of the x coordinates for the code to work correctly.

The code is

\documentclass{article}

\usepackage{tikz}
\usetikzlibrary{calc}

\begin{document}

\begin{tikzpicture}
\coordinate (1) at (0.1,0.2);
\coordinate (2) at (0.2,0.7);
\coordinate (3) at (0.4,-0.3);

\draw let \p1 = (1),
          \p2 = (2),
          \p3 = (3),
          \n{denom} = {(\x1 - \x2)*(\x1 - \x3)*(\x2-\x3)},
          \n{A} = {(\x3*(\y2-\y1) + \x2*(\y1-\y3) + \x1*(\y3-\y2))/\n{denom}},
          \n{B} = {(\x3*\x3*(\y1-\y2) + \x2*\x2*(\y3-\y1)+\x1*\x1*(\y2-\y3))/\n{denom}},
          \n{C} = {(\x2*\x3*(\x2-\x3)*\y1 + \x3*\x1*(\x3-\x1)*\y2 + \x1*\x2*(\x1-\x2)*\y3)/\n{denom}} in
          plot[domain=\x1:\x3] (\x,{\n{A}*\x*\x+\n{B}*\x + \n{C}});

\end{tikzpicture}

\end{document}

Answer 2

1) A variant with fp :

\documentclass{article}
\usepackage{tikz,fp}
\FPmessagesfalse
\FPdebugfalse

\makeatletter  
\tikzset{%
  parabola through/.style={
    to path={%
      \pgfextra{%
        \tikz@scan@one@point\pgfutil@firstofone(\tikztostart)\relax 
        \FPeval\xa{\pgf@sys@tonumber{\pgf@x}/28.45274}   
        \FPeval\ya{\pgf@sys@tonumber{\pgf@y}/28.45274}   
        \tikz@scan@one@point\pgfutil@firstofone#1\relax
        \FPeval\xb{\pgf@sys@tonumber{\pgf@x}/28.45274}   
        \FPeval\yb{\pgf@sys@tonumber{\pgf@y}/28.45274}   
        \tikz@scan@one@point\pgfutil@firstofone(\tikztotarget)\relax
        \FPeval\xc{\pgf@sys@tonumber{\pgf@x}/28.45274}   
        \FPeval\yc{\pgf@sys@tonumber{\pgf@y}/28.45274}   
        \FPeval\pb@a{(\ya*(\xb-\xc)+\yb*(\xc-\xa)+\yc*(\xa-\xb))/%
        ((\xa-\xb)*(\xa-\xc)*(\xb-\xc))}
        \FPeval\pb@b{(\ya*(\xc+\xb)*(\xc-\xb)+\yb*(\xa+\xc)*(\xa-\xc)+\yc*(\xb+\xa)*(\xb-\xa))/((\xa-\xb)*(\xa-\xc)*(\xb-\xc))} 
        \FPeval\pb@c{(\ya*\xb*\xc*(\xb-\xc)+\yb*\xa*\xc*(\xc-\xa)+\yc*\xa*\xb*(\xa-\xb))/((\xa-\xb)*(\xa-\xc)*(\xb-\xc))} 

    \draw plot[domain=\xa:\xc]  (\x,{\pb@a*(\x*\x)+\pb@b*\x+\pb@c}) ;
    }(\tikztotarget)
    }
  }
}
\makeatother
\begin{document}

\begin{tikzpicture}
\draw [help lines] (-3,-1) grid (7,4);  
\draw (-3,0) to[parabola through={(-2,2)}]%
  (0,-1) to[parabola through={(2,4)}] (4,0) to[parabola through={(5,3)}] (7,0);     
\end{tikzpicture}

\end{document} 

2) From maeshtro's answer with gnuplot

\documentclass{article}
\usepackage{tikz}
\makeatletter
\tikzset{%
  parabola through/.style={
    to path={%
      \pgfextra{%
        \tikz@scan@one@point\pgfutil@firstofone(\tikztostart)\relax 
        \edef\xa{\pgf@sys@tonumber{\pgf@x}}   
        \edef\ya{\pgf@sys@tonumber{\pgf@y}}   
        \tikz@scan@one@point\pgfutil@firstofone#1\relax
        \edef\xb{\pgf@sys@tonumber{\pgf@x}}   
        \edef\yb{\pgf@sys@tonumber{\pgf@y}}   
        \tikz@scan@one@point\pgfutil@firstofone(\tikztotarget)\relax
        \edef\xc{\pgf@sys@tonumber{\pgf@x}}   
        \edef\yc{\pgf@sys@tonumber{\pgf@y}}   

\draw plot[domain=\xa/28.45274:\xc/28.45274] function{
  \ya/28.45274*((x*28.45274-\xb)*(x*28.45274-\xc))/((\xa-\xb)*(\xa-\xc))+
  \yb/28.45274*((x*28.45274-\xa)*(x*28.45274-\xc))/((\xb-\xa)*(\xb-\xc))+
  \yc/28.45274*((x*28.45274-\xa)*(x*28.45274-\xb))/((\xc-\xa)*(\xc-\xb))
};
    }(\tikztotarget)
    }
  }
}
\makeatother
\begin{document}


\begin{tikzpicture}
    \draw [help lines] (-3,-1) grid (7,4);  
  \draw (-3,0) to[parabola through={(-2,2)}] (0,-1) to[parabola through={(2,4)}] (4,0) to[parabola through={(5,3)}] (7,0);     
\end{tikzpicture}

\end{document} 

Answer 3

Here is a solution inspired by this answer.

\documentclass{standalone}
\usepackage{tikz}
\makeatletter
\def\parabola@save@target#1{%
  \def\parabola@target{#1}}
\def\parabola@save@start#1{%
  \def\parabola@start{#1}}
\def\parabola@save@midpoint#1{%
  \def\parabola@midpoint{#1}}
\tikzset{
  parabola through/.style={
    to path={%
      \pgfextra{%
        \edef\parabola@@target{(\tikztotarget)}%
        \tikz@scan@one@point\parabola@save@target\parabola@@target\relax
        \edef\parabola@@start{(\tikztostart)}%
        \tikz@scan@one@point\parabola@save@start\parabola@@start\relax
        \edef\parabola@@midpoint{(#1)}%
        \tikz@scan@one@point\parabola@save@midpoint\parabola@@midpoint\relax
        \parabola@start
        \pgfmathsetmacro{\parabola@xa}{\the\pgf@x/1cm}
        \pgfmathsetmacro{\parabola@ya}{\the\pgf@y/1cm}
        \parabola@midpoint
        \pgfmathsetmacro{\parabola@xb}{\the\pgf@x/1cm}
        \pgfmathsetmacro{\parabola@yb}{\the\pgf@y/1cm}
        \parabola@target
        \pgfmathsetmacro{\parabola@xc}{\the\pgf@x/1cm}
        \pgfmathsetmacro{\parabola@yc}{\the\pgf@y/1cm}
        % f(x) = ax^2 + bx + c
        % a=-(-x1*y3+x3*y1+x2*y3+x1*y2-x2*y1-x3*y2)/(x1*x3^2-x2*x3^2+x2*x1^2-x3*x1^2+x3*x2^2-x1*x2^2)
        % b=(-x1^2*y3+x1^2*y2+y1*x3^2-y2*x3^2+x2^2*y3-y1*x2^2)/((x1-x2)*(-x1*x3+x1*x2+x3^2-x2*x3))
        % c=(x1^2*x2*y3-x1^2*x3*y2-x2^2*x1*y3+y2*x1*x3^2+x2^2*x3*y1-y1*x2*x3^2)/((x1-x2)*(-x1*x3+x1*x2+x3^2-x2*x3))
        \pgfmathsetmacro{\parabola@a}{-(-\parabola@xa*\parabola@yc+\parabola@xc*\parabola@ya+\parabola@xb*\parabola@yc+\parabola@xa*\parabola@yb-\parabola@xb*\parabola@ya-\parabola@xc*\parabola@yb)/(\parabola@xa*\parabola@xc^2-\parabola@xb*\parabola@xc^2+\parabola@xb*\parabola@xa^2-\parabola@xc*\parabola@xa^2+\parabola@xc*\parabola@xb^2-\parabola@xa*\parabola@xb^2)}
        \pgfmathsetmacro{\parabola@b}{(-\parabola@xa^2*\parabola@yc+\parabola@xa^2*\parabola@yb+\parabola@ya*\parabola@xc^2-\parabola@yb*\parabola@xc^2+\parabola@xb^2*\parabola@yc-\parabola@ya*\parabola@xb^2)/((\parabola@xa-\parabola@xb)*(-\parabola@xa*\parabola@xc+\parabola@xa*\parabola@xb+\parabola@xc^2-\parabola@xb*\parabola@xc))}
        \pgfmathsetmacro{\parabola@c}{(\parabola@xa^2*\parabola@xb*\parabola@yc-\parabola@xa^2*\parabola@xc*\parabola@yb-\parabola@xb^2*\parabola@xa*\parabola@yc+\parabola@yb*\parabola@xa*\parabola@xc^2+\parabola@xb^2*\parabola@xc*\parabola@ya-\parabola@ya*\parabola@xb*\parabola@xc^2)/((\parabola@xa-\parabola@xb)*(-\parabola@xa*\parabola@xc+\parabola@xa*\parabola@xb+\parabola@xc^2-\parabola@xb*\parabola@xc))}
        \draw plot[samples=100,domain=\parabola@xa:\parabola@xc] function {\parabola@a*(x**2)+\parabola@b*x+\parabola@c};
      }
    }
  }
}
\makeatother
\begin{document}
\begin{tikzpicture}
  \node[circle,fill=red] at (0,0) {};
  \node[circle,fill=blue] at (2,2) {};
  \node[circle,fill=green] at (4,0) {};
  \draw (0,0) to[parabola through={(2,2)}] (4,0);
\end{tikzpicture}
\end{document}

Answer 4

You can rely on \pgfmathparse to always return any given length in pt. Take a look at the output of:

\pgfmathparse{12cm+1pt}
\pgfmathresult

\pgfmathparse{1pt}
\pgfmathresult

which will output as:

342.43306
1.0

With this you can pretty much do all your calculation by transferring to the (now dimensionless) unit pt. However you should be weary of very large numbers.

To circumvent this you could rely on the fpu unit of pgf.

% Preamble
\usepgflibrary{fpu}

\begin{document}

\pgfkeys{/pgf/fpu,/pgf/fpu/output format=fixed}
\pgfmathparse{12cm+1pt}
\pgfmathresult

\pgfmathparse{1pt}
\pgfmathresult

342.43306000000000
1.0000000000

Then you have access to numbers which are always in one specific unit!

Answer 5

Thank you all for your answers. They helped me a lot. Finally I chose the following approach which uses gnuplot. It is not very elegant but it meets my needs.

\documentclass{standalone}
\usepackage{tikz}
\usetikzlibrary{calc}

\newcommand\parabola[3]{%
\path[draw]
let
  \p1=#1,\p2=#2,\p3=#3
in
\pgfextra
  \pgfmathsetmacro{\xa}{\x1/1cm}
  \pgfmathsetmacro{\ya}{\y1/1cm}
  \pgfmathsetmacro{\xb}{\x2/1cm}
  \pgfmathsetmacro{\yb}{\y2/1cm}
  \pgfmathsetmacro{\xc}{\x3/1cm}
  \pgfmathsetmacro{\yc}{\y3/1cm}
\endpgfextra
plot[domain=\xa:\xc] function{
  \ya*((x-\xb)*(x-\xc))/((\xa-\xb)*(\xa-\xc))+
  \yb*((x-\xa)*(x-\xc))/((\xb-\xa)*(\xb-\xc))+
  \yc*((x-\xa)*(x-\xb))/((\xc-\xa)*(\xc-\xb))
};}

\begin{document}
\begin{tikzpicture}
\coordinate (A) at (1,0);
\coordinate (B) at (3,8);
\coordinate (C) at (4,-1);
\fill[red] (A) circle (2pt) (B) circle (2pt) (C) circle (2pt); 
\parabola{(A)}{(B)}{(C)}  
\end{tikzpicture}
\end{document}