How to draw rotated parabola in LaTeX with using tikz and Bezier curve?

Question

I tried this code:

\begin{tikzpicture}
    \draw[semithick,->] (-2,0) -- (2,0) node[right] {$ x_1 $};
    \draw[semithick,->] (0,-2) -- (0,2) node[left] {$ x_2 $};   
    \path[thick,blue,draw] (-2,2) .. controls (0,1) .. (0,0) .. controls (0,-1) .. (2,-2);
    \path[thick,red,rotate=90,draw] (0,0) parabola (-2,-2);
\end{tikzpicture}

and got some curve (blue), but it's not a parabola!

What I've done wrong?

Answer 1

In general, a Bézier is a cubic (a polynomial at³+bt²+ct+d), but you only want a quadratic (bt²+ct+d so you want a=0).

To force a Bézier A .. controls B and C .. D to be a parabola, you make ABCD a trapezium with AD||BC and with AD=3BC. The point where the parabola is parallel to AD and BC is the intersection of the diagonals, and is 1/4 the way along the diagonals.

So if you make ABCD an isosceles trapezium then the axis of the parabola will be aligned with the axis of the trapezium and the vertex (marked "apex" below - oops) will be at this point of intersection of the diagonals.

\documentclass{standalone}
\usepackage{tikz}
\begin{document}
\begin{tikzpicture}
    \draw [help lines,black!20!white] (0,-3) grid (10,3);
    \draw[blue,dashed] 
        (0,0) coordinate(A)
        -- (3,-3) coordinate (B) 
        -- (4,-2) coordinate (C) 
        -- (3,3) coordinate (D) -- cycle
        (A)--(C) (B)--(D);
    \draw (A) .. controls (B) and (C) .. (D);
    \draw[red,dashed]
        (4,3) coordinate (P)
        -- (6,-1) coordinate(Q)
        -- (8,-1) coordinate(R)
        -- (10,3) coordinate(S) -- cycle (P)--(R) (Q)--(S);
    \draw (P) .. controls (Q) and (R) .. (S);
    \draw (7,0) circle (1mm) node[right,rotate=90]{apex};
    \foreach \lbl/\pos in {A/left,B/below,C/right,D/above,P/above,Q/below,R/below,S/above}
        \draw (\lbl) circle[radius=.5mm] (\lbl) node[\pos]{$\lbl$};
\end{tikzpicture}
\end{document}

FWIW the Bézier curve with identical control points — the curve that isn't a parabola in the original question — is a known curve. For example (1,0) .. controls (0,0) .. (0,1) and the equivalent code (1,0) .. controls (0,0) and (0,0) .. (0,1) give a segment of a superellipse with n=1/3

Answer 2

You could also parametrize the curve using

x(t)=t^2
y(t)=t

and then plot it using pgfplots

% arara: pdflatex
% !arara: indent: {overwrite: on}
\documentclass{standalone}
\usepackage{pgfplots}

% arrows as stealth fighters
\tikzset{>=stealth}
\begin{document}

\begin{tikzpicture}
    \begin{axis}[
            axis lines=middle,
            axis line style=<->,
            xmin=-5,xmax=5,
            ymin=-5,ymax=5,
            xlabel=$x_1$,
            ylabel=$x_2$,
            xtick=\empty,
            ytick=\empty,
            xticklabels=\empty,
            yticklabels=\empty,
        ]
        \addplot[smooth,thick, blue,-]({x^2},{x});
        \addplot[smooth,thick, red,-]({-x^2},{x});
    \end{axis}
\end{tikzpicture}

\end{document}

Answer 3

This is how Bezier curve is drawn. It requires 4 points where (x1,y1) and (x4,y4) are starting and end points while (x2,y2) nand (x3,y3) are auxiliary points, constituting a rectangle-like form and a continuous Bezier curve is plotted within. If only (x2,y2) point is given, it will be repeated for (x3,y3) and the curve will be sharpter, instead of flatter (see the plots below). So the location of (x2,y2) and (x3,y3) do affect the curvature.

(x1,y1) .. controls (x2,y2) and (x3,y3) .. (x4,y4);

The plot on the right is newly added for comparison.

Code

\documentclass[border=10pt]{standalone}
\usepackage{tikz}
\begin{document}

\begin{tikzpicture}
    \draw[semithick,->] (-2,0) -- (2,0) node[right] {$ x_1 $};
    \draw[semithick,->] (0,-2) -- (0,2) node[left]  {$ x_2 $};   
    \path[thick,blue,draw] (-2,2) .. controls (0,1) .. (0,0) .. controls (0,-1) .. (2,-2);
    \path[thick,red,rotate=90,draw] (0,0) parabola (-2,-2);

    \path[thick,blue,draw] (-2,2) .. controls (0,1) .. (0,0) .. controls (0,-1) .. (2,-2);
\end{tikzpicture}
\begin{tikzpicture}
    \draw[semithick,->] (-2,0) -- (2,0) node[right] {$ x_1 $};
    \draw[semithick,->] (0,-2) -- (0,2) node[left]  {$ x_2 $};   
    \path[thick,yellow,draw] (-2,2) .. controls (0.65,1) and (0.65,-1) .. (-2,-2) node[right]{curve 1};
    \path[thick,yellow,draw] (2,2) .. controls (-0.65,1) and (-0.65,-1) .. (2,-2)node[right]{curve 2};
    \path[thick,cyan,draw]   (2,2) node[right]{curve 3} .. controls (-0.65,0) .. (2,-2);
    \path[thick,cyan,draw]   (-2,2)node[right]{curve 4} .. controls (0.65,0) .. (-2,-2);    % curve 3 and 4 are sharper because (x2,y2)=(x3,y3)
\end{tikzpicture}
\end{document}