圓錐曲線的切線

Question

我有一個橢圓，我想在 A 點畫一條切線，但我不知道怎麼做。

\documentclass{standalone}
\RequirePackage{tikz}
\usetikzlibrary{calc,through,arrows,fadings,decorations.pathreplacing,matrix}

\begin{document}

\begin{center}
\begin{tikzpicture}[scale=1.4]
\clip (-3,-2) rectangle (3,2);
\coordinate (A) at (130:2 and 1);
\draw[black,thick] (0,0) ellipse (2 and 1);
\foreach \p in {A}
  \fill[black] (\p) circle (0.04);
\draw[black] ($(A)+(90:0.3)$) node{$A$};
\end{tikzpicture}
\end{center}

\end{document}

任何幫助將非常感激。

Answer 1

你的問題與以下密切相關這個，我稍微修改了那裡的答案以獲得

\documentclass[tikz,border=3.14mm]{standalone}
\usepackage{pgfplots}
\usepgfplotslibrary{fillbetween}
\usetikzlibrary{calc,decorations.markings}

\begin{document}
\begin{tikzpicture}[scale=1.4,tangent at/.style={% cf. https://tex.stackexchange.com/questions/25928/how-to-draw-tangent-line-of-an-arbitrary-point-on-a-path-in-tikz/25940#25940
    decoration={ markings,
      mark =at position #1 with {\draw[purple,-latex](-1,0) -- (1,0);},
    }, decorate
  }]
\clip (-3,-2) rectangle (3,2);
\coordinate (A) at (130:2 and 1);
\draw[black,thick,name path=elli] (0,0) ellipse (2 and 1);
\node[label=above:$A$,circle,fill,inner sep=0.05cm] at (A){};
\path[name path=clip] (130:1 and 0.5) -- (130:2.2 and 1.1);
\path [%draw,blue,
    name path=middle arc,
    intersection segments={
        of=elli and clip,
        sequence={A1}
    },
    postaction={tangent at/.list={0}}]; 
\end{tikzpicture}
\end{document}

這是適用於您不知道參數化的曲線的通用方法。

當然，對於橢圓來說，這有點過分了。您只需分析計算正切即可。

\documentclass[tikz,border=3.14mm]{standalone}
\begin{document}
\begin{tikzpicture}[scale=1.4]
\clip (-3,-2) rectangle (3,2);
\coordinate (A) at (130:2 and 1);
\draw[black,thick] (0,0) ellipse (2 and 1);
\node[label=above:$A$,circle,fill,inner sep=0.05cm] at (A){};
\draw[blue] (A) -- ++ ({-2*sin(130)},{cos(130)})
(A) -- ++ ({2*sin(130)},{-cos(130)});
\node[above]{$\gamma(t)=\bigl(2\,\cos(t),\sin(t)\bigr)$};
\node[below]{$\dot\gamma(t)=\bigl(-2\,\sin(t),\cos(t)\bigr)$};
\end{tikzpicture}
\end{document}

Answer 2

2022 年更新Asymptote 更輕鬆！

// http://asymptote.ualberta.ca/
unitsize(1cm);
path p=ellipse((0,0),3,1.8);
draw(p);
real t=1.5;
pair M=point(p,t);
pair Mt=dir(p,t);        // the tangent vector
pair Mn=rotate(90)*Mt;   // the normal vector
draw(M-3Mt--M+2Mt,red);
draw(M-Mn--M+Mn,blue);

---

舊答案這個geometric答案比analytic一個還長。我喜歡用幾何方法來構造橢圓上任意點的切線。切線垂直於點視圖到 2 個焦點的角度的平分線。

\documentclass[tikz,border=5mm]{standalone}
\usetikzlibrary{calc}
% Usage:\bisectorpoint(A,B,C)(D)
% >>> D is on AC such that  
% BD is perpendicular bisector of the angle ABC
\def\bisectorpoint(#1,#2,#3)(#4){% in the PGF manual
\path let
\p1=($(#2)-(#1)$),\p2=($(#2)-(#3)$),
\n1={veclen(\x1,\y1)},\n2={veclen(\x2,\y2)}
in ($(#1)!scalar(\n1/(\n1+\n2))!(#3)$) coordinate (#4);
}%
\begin{document}
\begin{tikzpicture}
\def\a{4}     % major of ellipse
\def\b{2.5}   % minor of ellipse
\pgfmathsetmacro{\c}{sqrt(\a*\a-\b*\b)}
\path
(-\c,0) coordinate (F1) node[below]{$F_1$}
(\c,0)  coordinate (F2) node[below]{$F_2$}
(140:{\a} and {\b}) coordinate (A)
;
\bisectorpoint(F1,A,F2)(P)
\path 
(P)--(A)--([turn]90:4)  coordinate (A1)
(P)--(A)--([turn]-90:3) coordinate (A2);
\begin{scope} 
\clip (P)--(A)--(F1);
\draw[gray] (A) circle(8mm);
\end{scope}
\begin{scope} 
\clip (P)--(A)--(F2);
\draw[gray] (A) circle(6mm);
\end{scope}

\draw[gray,shorten <=-1cm] (A)--(P) node[below]{$P$}; 
\draw (F1)--(A)--(F2) 
(-\a-.5,0)--(\a+.5,0) (0,-\b-.5)--(0,\b+.5); 
\draw circle({\a} and {\b});
\draw[magenta] (A1)--(A2);
\end{tikzpicture}   
\end{document}

Answer 3

使用tzplot:

\documentclass{standalone} 

\usepackage{tzplot}

\begin{document}

\begin{tikzpicture}[scale=1.4]

\tzellipse[thick]"AA"(0,0)(2 and 1)
\tzcoor(130:2 and 1)(A){$A$}
\tztangent{AA}(A)[-2.5:0]

\end{tikzpicture}

\end{document}