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Tikz：不使用角度绘制圆弧

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Tikz：不使用角度绘制圆弧

是否可以不使用角度而是使用两个位置来绘制圆弧？

\documentclass{article}
\usepackage{tikz}
\begin{document}
\begin{tikzpicture}
  \pgfmathsetmacro{\a}{2.5}
  \pgfmathsetmacro{\b}{2}
  \pgfmathsetmacro{\c}{sqrt(\a^2 - \b^2)}
  \draw (0, 0) ellipse (\a cm and \b cm);
  \filldraw[black] (\c, 0) circle (.05cm) node[right = 2pt, scale = .75]
  {\(F\)};
  \begin{scope}[decoration = {markings,
      mark = at position 0.5 with {\node[circle, inner sep = .01cm,
        fill = white, scale = .75] {\(\mathbf{r}_1\)};},
    }]
    \draw[postaction = decorate]
    (\c , 0) -- ({\a * cos(90)}, {\b * sin(90)});
  \end{scope}
  \begin{scope}[decoration = {markings,
      mark = at position 0.5 with {\node[circle, inner sep = .01cm,
        fill = white, scale = .75] {\(\mathbf{r}_2\)};},
    }]
    \draw[postaction = decorate]
    (\c , 0) -- ({\a * cos(60)}, {-\b * sin(60)});
  \end{scope}
  \begin{scope}[decoration = {markings,
      mark = at position 0.5 with {\node[circle, inner sep = .01cm,
        fill = white, scale = .75] {\(2a - r_1\)};},
    }]
    \draw[postaction = decorate]
    (-\c , 0) -- ({\a * cos(90)}, {\b * sin(90)});
  \end{scope}
  \begin{scope}[decoration = {markings,
      mark = at position 0.5 with {\node[circle, inner sep = .01cm,
        fill = white, scale = .75] {\(2a - r_2\)};},
    }]
    \draw[postaction = decorate]
    (-\c , 0) -- ({\a * cos(60)}, {-\b * sin(60)});
  \end{scope}
  \filldraw[draw = black, fill = white] (-\c, 0) circle (.05cm)
  node[left = 2pt, scale = .75]
  {\(F^*\)};
  \filldraw[black] ({\a * cos(60)}, {-\b * sin(60)})
  node[below = 2pt, scale = .75] {\(P_2\)};
  \filldraw[black] ({\a * cos(90)}, {\b * sin(90)})
  node[above = 2pt, scale = .75] {\(P_1\)};
\end{tikzpicture}}
\end{document}

我想从 到 画一条弧线，({\a * cos(60) / 4}, {-\b * sin(60) / 4})而不({\a * cos(90) / 4}, {\b * sin(90) / 4})需要计算角度（这很简单，但我想让乳胶来完成这项工作）。

这该怎么做？如果做不到，那我就自己算一下，自己完成。

在此处输入图片描述

答案1

您可以使用以下方法之一使用 tikz 标签角度。这里，我将 Shiyu 的答案包装成一个宏，该宏以中心点和射线上的两个点为参数，绘制圆弧。它还接受一个可选参数，可用于设置选项draw。

\def\angleRadius{0.5cm}
\newcommand{\markangle}[4][]{
    \draw[#1] let \p1=#2, \p2=#3, \p3=#4, 
         \n1={atan2(\x2-\x1,\y2-\y1)}, \n2={atan2(\x3-\x1,\y3-\y1)} in
         ($(\p1)!\angleRadius!(\p3)$) arc [start angle=\n2-360, end angle=\n1, radius=\angleRadius]; 
}

我稍微简化了你的代码，你不需要所有这些scope和markings装饰，只需将节点放在线上：

\documentclass[border=5mm]{standalone}
\usepackage{tikz}
\usetikzlibrary{calc}
\def\angleRadius{0.5cm}
\newcommand{\markangle}[4][]{
    \draw[#1] let \p1=#2, \p2=#3, \p3=#4, 
         \n1={atan2(\x2-\x1,\y2-\y1)}, \n2={atan2(\x3-\x1,\y3-\y1)} in
         ($(\p1)!\angleRadius!(\p3)$) arc [start angle=\n2-360, end angle=\n1, radius=\angleRadius]; 
}


\begin{document}
\begin{tikzpicture}[
    radius=0.05cm
]
  \pgfmathsetmacro{\a}{2.5}
  \pgfmathsetmacro{\b}{2}
  \pgfmathsetmacro{\c}{sqrt(\a^2 - \b^2)}
  \draw (0, 0) ellipse [x radius=\a cm, y radius=\b cm];
  \filldraw[black] (\c, 0) circle node[right = 2pt, scale = .75]
  {$F$};
  \draw (\c,0) -- ({\a * cos(90)}, {\b * sin(90)})
    node [pos=0.5, fill=white, font=\small, inner sep=2pt] {$\mathbf{r}_1$};
  \draw (\c , 0) -- ({\a * cos(60)}, {-\b * sin(60)})
            node [pos=0.5, fill=white, font=\small, inner sep=2pt] {$\mathbf{r}_2$};;
  \draw (-\c , 0) -- ({\a * cos(90)}, {\b * sin(90)})
            node [pos=0.5, fill=white, font=\small, inner sep=2pt] {$2a - r_1$};;
  \draw (-\c , 0) -- ({\a * cos(60)}, {-\b * sin(60)})
            node [pos=0.5, fill=white, font=\small, inner sep=2pt] {$2a - r_2$};;
  \filldraw[draw = black, fill = white] (-\c, 0) circle
  node[left, font=\small] {\(F^*\)};
  \node at ({\a * cos(60)}, {-\b * sin(60)}) [below = 2pt, scale = .75] {\(P_2\)};
  \node at ({\a * cos(90)}, {\b * sin(90)}) [above = 2pt, scale = .75] {\(P_1\)};
  \markangle[thick, red]{(\c, 0)}{({\a * cos(60)}, {-\b * sin(60)})}{({\a * cos(90)}, {\b * sin(90)})}
\end{tikzpicture}
\end{document}

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