感谢群友@香港-小狮子
提出了一种较为简单的几何画法!其余参与讨论的大家,一并感谢您们的指导!
方法肯定不止这一种,后续我将继续整理讨论结果,同时十分欢迎提出新的回答!
\documentclass[tikz,border=2pt]{standalone}
\usetikzlibrary{calc,intersections}
\usepackage{amsmath}
\begin{document}
\begin{tikzpicture}[scale=2]
\draw [->] (-3,0) -- (3,0) node [below left] {$x$};
\draw [->] (0,-3) -- (0,3) node [below left] {$y$};
\draw [help lines] (-3,-3) grid (3,3);
\clip (-3.1,-3.1) rectangle (3.1,3.1);
\draw [thick] (2,0) arc (0:360:2 and 1) node at (-2,-2) {$\dfrac{x^2}{4}+y^2=1$};
\node (O) at (0,0) [below left] {$O$};
\fill (1,2) circle (1pt) coordinate (P) node [above right] {$P(1,2)$};
\fill ({-sqrt(3)*1cm},0cm) circle (1pt) coordinate (F1) node [below] {$F_1$};
\fill ({sqrt(3)*1cm},0cm) circle (1pt) coordinate (F2) node [below] {$F_2$};
\path let \p1=($(P)-(F2)$),\n1={veclen(\x1,\y1)}
in [dotted,name path=cirP] (P) circle (\n1);
\path [dotted,name path=cirF1] (F1) circle (4);
\path [name intersections={of=cirP and cirF1,by={J1,J2}},dashed] (J1)--(F2) (J2)--(F2);
\draw let \p1=(J1),\p2=(F2),\n1={\y2-\y1},\n2={\x2-\x1},\n3={atan{\n1/\n2}-90}
in [ultra thick,red] (P)--++(\n3:4cm);
\draw let \p1=(J2),\p2=(F2),\n1={\y2-\y1},\n2={\x2-\x1},\n3={atan{\n1/\n2}-90}
in [ultra thick,red] (P)--++(\n3:3cm);
\end{tikzpicture}
\end{document}
问 过标准椭圆外一定点,引椭圆的两条切线,绘制切线