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隐含的情节——省略号

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隐含的情节——省略号

我需要画一个圆锥曲线：给定两个椭圆

$\left( \frac{x}{a_1}\right)^2 + \left( \frac{y}{b_1} \right)^2 - 1 = 0$

和

$\left( \frac{x}{a_2}\right)^2 + \left( \frac{y}{b_2} \right)^2 - 1 = 0$,

我要密谋

$$
t_1 \left( \left( \frac{x}{a_1}\right)^2 + \left( \frac{y}{b_1} \right)^2 - 1 \right) + t_1 \left( \left( \frac{x}{a_2}\right)^2 + \left( \frac{y}{b_b} \right)^2 - 1 \right) = 0
$$

对于$t_1$和的某些值$t_2$。

我理解 Ti钾Z 可能不适合绘制隐函数。你有什么在 Ti 中执行此操作的技巧吗钾Z 或其他程序的建议？

答案1

这是一个使用选项pgfplots和gnuplot（带--shell-escape选项的处理）：

\documentclass{article}
\usepackage{pgfplots}

\pgfplotsset{compat=newest}

\begin{document}

\begin{tikzpicture}
  \begin{axis}
    \addplot +[no markers,
      raw gnuplot,
      thick,
      ] gnuplot {
      set contour base;
      set cntrparam levels discrete 0.003;
      unset surface;
      set view map;
      set isosamples 500;
      splot (x/2)^2+ (y/3)^2 -1;
    };
    \addplot +[no markers,
      red,
      raw gnuplot,
      thick,
      ] gnuplot {
      set contour base;
      set cntrparam levels discrete 0.003;
      unset surface;
      set view map;
      set isosamples 500;
      splot (x/5)^2+ (y/2)^2 -1;
    };
\foreach \Valora/\Valorb in {0/1,0.25/0.75,0.5/0.5,0.75/0.25,1/0}
{
    \addplot +[no markers,
      raw gnuplot,
      thin,
      dashed
      ] gnuplot {
      set contour base;
      set cntrparam levels discrete 0.003;
      unset surface;
      set view map;
      set isosamples 500;
      splot \Valora*((x/2)^2+ (y/3)^2 -1) + \Valorb*((x/5)^2+ (y/2)^2 -1);
    };
}
\end{axis}
\end{tikzpicture}

\end{document}

在此处输入图片描述

答案2

如果您真的想绘制圆锥束，Asymptote 可以用更几何的方式完成此操作。它有一个例程来计算通过任意给定五个点（没有三个共线）的圆锥，因此您实际上可以通过指定的四个点的基轨迹来计算圆锥束。

[数学注释：如果一条线恰好经过四个点中的一个，则存在自然同构（线上的点）->（铅笔中的圆锥曲线），将一个点引向与该点和四个固定点相交的唯一圆锥曲线。]

圆锥曲线

% file: foo.tex
% to compile: pdflatex --shell-escape foo
%
% For MikTeX users: Asymptote requires a separate program that cannot be installed
% by the package manager. You can get the installation file from
% https://sourceforge.net/projects/asymptote/files/2.35/
% (specifically, the file ending in setup.exe).
\documentclass{standalone}
\usepackage{asypictureB}
\begin{document}
\begin{asypicture}{name=PencilOfConics}
settings.outformat = "pdf";
import geometry;  // Import the geometry module to use conics
unitsize(1cm);
pair a=(-3,-2), b=(-3,2), c=(2.5,1.5), d=(3,-2);
pair fifthpoint(real t) { return d + (0,t); }
real tmin = -3-.01*unitrand(), tmax = 3 + 0.01*unitrand();
int n = 20;
pair[] points = new pair[n];
for (int i = 0; i <= n; ++i) {
  real t = interp(tmin, tmax, i/n);
  pair pt = fifthpoint(t);
  draw(conic(a,b,c,d,pt), interp(blue, red, i/n));
  points[i] = pt;
}
dot(points);
\end{asypicture}
\end{document}

答案3

运行xelatex：

\documentclass{article}
\usepackage{pst-func}
\begin{document}
\begin{pspicture*}(-5.5,-5.5)(5.5,5.5)
\psaxes{->}(0,0)(-5,-5)(5,5)
\def\aI{2}  \def \aII{3}
\def\bI{-1} \def \bII{-2}
\pgfforeach \tI/\tII in {0/1,0.25/0.75,0.5/0.5,0.75/0.25,1/0}{%
  \psplotImp[algebraic,linewidth=1.5pt,linecolor=red](-6,-6)(6,6)
    {\tI *((x/\aI)^2 +(y/\bI)^2-1) + \tII*((x/\aII)^2+(y/\bII)^2-1)}}
\end{pspicture*}
\end{document}

在此处输入图片描述

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