How to plot a Hyperbolic paraboloid in a style of the attached figure

Question

How to plot a Hyperbolic paraboloid in a style of the attached figure using only one COLOUR

Modify my Code please

   \documentclass{article}
\usepackage{pgfplots}
\begin{document}
\begin{figure}
    \centering
    \begin{tikzpicture}
        \begin{axis}[
            xlabel=$x$,
            ylabel=$y$,
            zlabel=$z$,
            view={60}{30},
        ]
        \addplot3[surf, shader=interp, domain=-2:2, samples=41] 
            {x^2/4 - y^2/9};
        \end{axis}
    \end{tikzpicture}
    \caption{Hyperbolic Paraboloid: $z = \frac{x^2}{4} - \frac{y^2}{9}$}
\end{figure}
\end{document}

I think that addplot3 with surf draws the graph of a function defined on a rectangle. In your attached image, the surface is such a graph but with the Oz axis clipped. I don't know if this is possible with addplot3. In case it is not and you really need that type of representation, I guess you have to do it directly in TikZ where you control the domain of the parametrization (for example the y domain will depend of each x in the x domain). — Daniel N, Nov 14 '23 at 05:51

Daniel N · Accepted Answer · 2023-11-16T05:44:18.830

One solution would be to split the paraboloid in two, to achieve some sort of layering. I did the same thing with the coordinate axes.

I modified the code according to @hpekristiansen kind remark concerning the compatibility of the coordinate systems of pdfplots and tikz.

The code

\documentclass[11pt, margin=10pt]{standalone}
\usepackage{pgfplots}
\pgfplotsset{compat=1.17}
\begin{document}
\begin{tikzpicture}
  \pgfplotsset{set layers}
  \begin{axis}view={150}{12}, axis lines=none]
    \addplot3[
    surf,
    colormap/cool,
    shader=flat,
    opacity=.6,
    domain=-4:4,
    domain y=-4:0,
    samples=40,
    clip=false
    ] {x^2 -y^2};
% plane section x = 0
\addplot3[%
draw=blue, very thin,
fill=red, fill opacity=.15,
domain=-4:4,
samples y=0
] (0, x, {-x^2});

\draw[ultra thin, -&gt;] (0, 0, 0) -- (6, 0, 0)
node[below, text=black, scale=.8] {$x$};

\addplot3[
surf,
colormap/cool,
shader=flat,
opacity=.6,
domain=-4:4,
domain y=0:4,
samples=40,
clip=false
] {x^2 -y^2};

\addplot3[%
draw=blue, very thin,
domain=0:4,
samples y=0
] (0, x, {-x^2});

\draw[ultra thin, -&gt;] (0, 0, 0) -- (0, 0, 18)
node[above, text=black, scale=.8] {$z$};
\draw[ultra thin, -&gt;] (0, 0, 0) -- (0, 6, 0)
node[below, text=black, scale=.8] {$y$};

\end{axis}
\end{tikzpicture}
\end{document}

+1 Remember to always set the compat level when using PGFPlots. Do not use explicit axis cs: -it is the default coordinate system since many versions ago. — hpekristiansen, Nov 15 '23 at 09:09
Thank you @hpekristiansen I shall try your indications tonight to see if I understand them correctly. If I use, say \pgfplotsset{compat=1.??}, then I can write \draw (0, 0, 0) ... instead of \draw (axis cs: 0, 0, 0) ..., isn't it? — Daniel N, Nov 15 '23 at 10:31
Yes - exactly. Use preferably 1.18 if your system is up to date or a version not too old. If you do not you get a warning when you compile. See https://tex.stackexchange.com/a/139695/8650 — hpekristiansen, Nov 15 '23 at 10:52

score 0 · Answer 2 · answered Oct 13 '23 at 13:45

0

The image you show does not match the hyperboloid calculated.

See section 5.2 (page 425) of the pgfplots manual.

\documentclass{article}
\usepackage{pgfplots}
\usepgfplotslibrary{colorbrewer}
\begin{document}
\begin{figure}
    \centering
    \begin{tikzpicture}
        \begin{axis}[
            xlabel=$x$,
            ylabel=$y$,
            zlabel=$z$,
            view={60}{30},
            colormap/PuBu-9
            ]
        \addplot3[surf, shader=interp, domain=-2:2, samples=41] 
            {x^2/4 - y^2/9};
        \end{axis}
    \end{tikzpicture}
    \caption{Hyperbolic Paraboloid: $z = \frac{x^2}{4} - \frac{y^2}{9}$}
\end{figure}
\end{document}

answered Oct 13 '23 at 13:45

John Kormylo

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Thank you. I want to reproduce the above Fig that I have attached. It will help me to plot other Figs of conicoids or quadrics. Please help... – math131 Oct 13 '23 at 17:33
You can do a lot by changing the viewing angles, and reversing the signs. The shown image is not using an orthogonal 3D grid either. (the y axis should be sloped, or the x axis should point directly to the POV). – John Kormylo Oct 13 '23 at 19:41

How to plot a Hyperbolic paraboloid in a style of the attached figure

2 Answers2