[Tex/LaTex] Line tangent to a a curve (using plot or controls), starting at a point

tikz-pgf

I need to draw lines that are parallel to a curve starting at a point.

The curve could be a curve drawn using 3 point as my example below did, but alternatively could be a curve drawn using the \draw () .. controls () and () .. () command. Preferably it would work either way.

This is the picture of what I am after:

enter image description here

Explanation

I have to curves, u_0 and u_1, and a point I_0. I want to find the line that goes from this point (I_0) and is tangent to the first curve (u_0), and the line that goes from (I_0) and is tangent to the second curve (u_1).

Then, I want to find a line that is parallel to the line tangent to (u_1), and this line needs to be tangent to (u_0). If the line can start and stop at the axis without using the shorten commands even better.

For now, I found the tangents and the start and finish points of the parallel line with trial and error, but I have dozens of such pictures to drawn and I need a better way of doing this.

Here is my code:

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

\begin{document}

\begin{tikzpicture}
    % axes
    \coordinate (origin) at (0,0);
    \draw (origin) -- (0,6.5) node[left]{$x_{2}$};
    \draw (origin) -- (9.5,0) node[below]{$x_{1}$};
    % u_0
    \node (e) at (2.2,4.9) {};
    \node (f) at (3,3) {};
    \node (g) at (5,2.3) {};
    \draw[thick] plot[smooth,tension=0.9] coordinates {(e) (f) (g)} node[right]{$u_{0}$};
    % u_1
    \node (h) at (1.5,4.2) {};
    \node (b) at (2.4,2.3) {};
    \node (j) at (4.4,1.6) {};
    \draw[thick] plot[smooth,tension=0.9] coordinates {(h) (b) (j)} node[right]{$u_{1}$};
    % first budget line (I_0 to p_0)
    \coordinate[label=left:{\scriptsize$I_{0}$}] (i0) at (0,4.65);
    \draw (i0) -- (8.1,0); 
    % second budget line (I_0 to p_1)
    \coordinate (d) at (4.67,0) {};
    \draw (i0) -- (d);
    % dashed line
    \draw[densely dashed,style={shorten >=2.8cm,shorten <=-4.64cm}] (2.68,3.3)   -- +($(i0)-(d)$);
\end{tikzpicture}
\end{document}

Best Answer

I think a decoration idea can be utilized here. As can be seen in the manual, the decoration declaration has a particular useful register that keeps the decoration angle defined as the tangent at the particular decoration segment. So I played around with that idea and here is the result:

\documentclass{article}
\usepackage{tikz}
\usetikzlibrary{decorations}
\newcounter{loopcoun}
\setcounter{loopcoun}{0}
\makeatletter
\pgfdeclaredecoration{findtan}{initial}{
\state{initial}[width=1mm,next state=findit]{
}
\state{findit}[width=1mm]{
\ifnum\value{loopcoun}<1
        \pgfpointanchor{i0}{center}
        \pgf@xa = \pgf@x
        \pgf@ya = \pgf@y
        \pgfmathparse{ifthenelse(atan2(\pgf@xa,\pgf@ya)<\pgfdecoratedangle,1,}
        \ifx\pgfmathresult\@empty\relax
        \pgfmoveto{\pgfpointorigin}
        \pgflineto{\pgfpointanchor{i0}{center}}
        \setcounter{loopcoun}{1}
        \else
        \fi
\else\fi
\pgfusepath{stroke}
    }
\state{final}[1mm]{
\setcounter{loopcoun}{0}
}
}
\makeatother
\begin{document}
\begin{tikzpicture}
\draw[style=help lines] (0,0) grid[step=1cm] (8cm,5cm);
\node[fill,inner sep=1pt,circle] (i0) at (0,3) {};

\path[draw,decoration=findtan,postaction={decorate}] (2,3) .. controls (0,1) and (4,0) .. (5,1);
\path[red,draw,decoration=findtan,postaction={decorate}] (3,4) .. controls (2,1) and (5,2) .. (6,1);
\end{tikzpicture}
\end{document}

enter image description here

The basic idea is that TikZ starts travelling on the curve little by little (given by the width option of the \state declaration) and do some stuff defined by the commands of that particular state. Basically, that defines the approximation resolution (reduce if necessary!). Here I am calculating the angle of the tangent line and comparing it with the given (i0) point. If it satisfies the constraint I am drawing a line to the point and incrementing the counter such that the code gets executed only once. So this is a proof of the concept. If you don't increment the counter it would give all the points satisfying the condition. Note that we are really exploiting something that is not meant to behave like this at all so expect all kinds of strange results. At least it doesn't draw anything if there is no such point, please play around with the location of (i0) to see if there is any bugs.

I wrote up a slightly messy code that seemingly does what you wanted. The idea is that you supply some initial points and then draw the curves indicating the particular tangent point. Then if any, they will be collected as nodes under the generic name (c-#). With this I did some tedious calculations using some geometry and draw the lines. There is much to improve obviously and hope it helps.

edit: the curves should start from top left to bottom right since the code marks the first eligible point. first of possibly many shortcomings...

\documentclass{article}
\usepackage{tikz}
\usetikzlibrary{decorations,calc}
\newcounter{loopcoun}
\newcounter{cocounter}
\setcounter{cocounter}{0}
\setcounter{loopcoun}{0}
\makeatletter

\pgfkeys{/tikz/find tangent/.style={decoration=findtan,depoint=#1,postaction=decorate}}
\pgfkeys{/tikz/depoint/.code= \edef\pgf@pointname{#1}}

\pgfdeclaredecoration{findtan}{initial}{
\state{initial}[width=1mm,next state=findit]{
}
\state{findit}[width=1mm]{
\ifnum\value{loopcoun}<1
        \pgfpointanchor{\pgf@pointname}{center}
        \pgf@xa = \pgf@x
        \pgf@ya = \pgf@y
        \pgfmathparse{ifthenelse(atan2(\pgf@xa,\pgf@ya)>\pgfdecoratedangle,1,0)}
        \ifnum\pgfmathresult>0
        \stepcounter{cocounter}{1}
        \pgfcoordinate{c-\the\value{cocounter}}{\pgfpointorigin}
        \setcounter{loopcoun}{1}
        \else
        \fi
\else\fi
\pgfusepath{stroke}
    }
\state{final}[1mm]{
\setcounter{loopcoun}{0}
}
}
\makeatother
\begin{document}
\begin{tikzpicture}
\draw[style=help lines] (0,0) grid[step=1cm] (9cm,5cm);
\node[fill,inner sep=1pt,circle] (p0) at (0,3) {};
\node[fill,inner sep=1pt,circle] (p1) at (0,4) {};

\draw[find tangent=p0] (2,3) .. controls (0,1) and (4,0) .. (5,1);
\path[find tangent=p0] (3,4) .. controls (2,1) and (5,2) .. (6,1);
\draw[red,find tangent=p1] (3,4) .. controls (2,1) and (5,2) .. (6,1);

\draw let \p1 = (p0), \p2 = (c-1),\n2 = {atan2(\x1-\x2,\y1-\y2)},\n3 = {-\y1/sin(\n2)} in (p0) -- ++(\n2:\n3); 
\draw let \p1 = (p0), \p2 = (c-2),\n2 = {atan2(\x1-\x2,\y1-\y2)},\n3 = {-\y1/sin(\n2)} in (p0) -- ++(\n2:\n3); 
\draw[dashed] let \p1 = (p1), \p2 = (c-3),\n2 = {atan2(\x1-\x2,\y1-\y2)},\n3 = {-\y1/sin(\n2)} in (p1) -- ++(\n2:\n3); 
\end{tikzpicture}
\end{document}

enter image description here

Related Solutions

[Tex/LaTex] How to draw tangent line of an arbitrary point on a path in TikZ

Here's an approach that uses the decorations.markings library to place support coordinates at a specified distance along a path, which can be used to set a local coordinate system at a later time to draw tangents or orthogonal lines.

You specify a tangent point by using tangent=<pos> in your first path. In a later path, you can then set use tangent to set a local coordinate system for that path: (0,0) is the tangent point itself, (1,0) is 1 unit along the tangent line, (0,1) is one unit along the orthogonal line.

You can specify tangent=<pos> multiple times in your original path. You can then specify which tangent point to use for a new path by calling use tangent=<count>, where the tangent points are numbered in the order they were specified in the original path.

Using these styles, the following code

\draw [
    tangent=0.4,
    tangent=0.56
] (0,0)
    to [out=20,in=120] (5,2)
    to [out=-60, in=110] (9,1);
\draw [blue, thick, use tangent] (-3,0) -- (3,0);
\draw [orange, thick, use tangent=2] (-2,0) -- (2,0) (0,0) -- (0,1);

would yield this

Here's the complete code:

\documentclass{article}
\usepackage{tikz}
\usetikzlibrary{decorations.markings}

\begin{document}

\begin{tikzpicture}[
    tangent/.style={
        decoration={
            markings,% switch on markings
            mark=
                at position #1
                with
                {
                    \coordinate (tangent point-\pgfkeysvalueof{/pgf/decoration/mark info/sequence number}) at (0pt,0pt);
                    \coordinate (tangent unit vector-\pgfkeysvalueof{/pgf/decoration/mark info/sequence number}) at (1,0pt);
                    \coordinate (tangent orthogonal unit vector-\pgfkeysvalueof{/pgf/decoration/mark info/sequence number}) at (0pt,1);
                }
        },
        postaction=decorate
    },
    use tangent/.style={
        shift=(tangent point-#1),
        x=(tangent unit vector-#1),
        y=(tangent orthogonal unit vector-#1)
    },
    use tangent/.default=1
]
\draw [
    tangent=0.4,
    tangent=0.56
] (0,0)
    to [out=20,in=120] (5,2)
    to [out=-60, in=110] (9,1);
\draw [blue, thick, use tangent] (-3,0) -- (3,0);
\draw [orange, thick, use tangent=2] (-2,0) -- (2,0) (0,0) -- (0,1);
\end{tikzpicture}
\end{document}

[Tex/LaTex] Tangent to circle given point on circle

The voodoo effect comes from the fact that when a coordinate is omitted origin is assumed and this happens to be the center of the circle. If you change the starting point the mystery goes away rather quickly.

I've placed more arrows to show the effect when the paths start from different coordinates.

\documentclass[tikz]{standalone}
\usetikzlibrary{calc}
\def\rad{2cm}

\begin{document}
\begin{tikzpicture}
 [point/.style = {draw, circle, fill=black, inner sep=0.5pt}]
\draw[style=help lines] (0,-1) grid[step=1cm] (5,4);

\node (C) at (2,1) [point,label=0:C]{};
\draw (C) circle (\rad);
\path (2,1) node[point,label={180:P}] (P)  at +(120:\rad){};

\foreach \x in {0,10,...,90}{
\draw[-latex,draw=blue,thick] (2,1) -- (P) -- ([turn]\x:2cm);
\draw[-latex,draw=red] (P) -- ([turn]\x:2cm);% You can add (0,0) -- as an initial point too
}
\end{tikzpicture}
\end{document}

enter image description here

As you can see, when initiated from a different point the red arrows loose the magic tangentiality but rather follow the incoming angle to that point (though blue arrows still preserve since the initial point is given). When the initial point is omitted it assumes that the path starts from (0,0) hence there is an inherent illusion of guessing the tangent.

Best Answer

Related Solutions

[Tex/LaTex] How to draw tangent line of an arbitrary point on a path in TikZ

[Tex/LaTex] Tangent to circle given point on circle

Related Question