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试试tkz-euclide宏包
转化成eps
一直没有注意到这个问题，今天用tkz-euclide凑了一个(椭圆)：
\documentclass[margin=3pt]{standalone}
\usepackage{tkz-euclide}
\begin{document}
\begin{tikzpicture}
\tkzDefPoint(-2,0){O}
\tkzDefPoint(1,6){J}
\tkzDefMidPoint(O,J)
\tkzGetPoint{I}
\tkzInterCC(J,I)(I,J) \tkzGetPoints{L}{H}
\tkzInterCC(O,I)(I,O) \tkzGetPoints{K}{G}
\tkzInterLL(O,K)(J,H) \tkzGetPoint{M}
\tkzInterLL(O,G)(J,L) \tkzGetPoint{N}
\tkzDefPointsBy[symmetry=center J](L,H){D,E}
\tkzDefPointsBy[symmetry=center O](G,K){C,F}
\tkzDefMidPoint(O,C)
\tkzGetPoint{P}
\tkzDefPointOnCircle[through = center J angle 30 point E]
\tkzGetPoint{A}
\tkzDefLine[tangent at=A](J)
\tkzGetPoint{h}
\tkzDefLine[orthogonal=through A](h,A) \tkzGetPoint{a}
\tkzDefLine[orthogonal=through O](A,h) \tkzGetPoint{b}
\tkzInterLL(O,b)(A,h) \tkzGetPoint{B}
\tkzDefPointOnLine[pos=0.15](O,A)\tkzGetPoint{A'}
\tkzDefPointOnLine[pos=0.15](O,B)\tkzGetPoint{B'}
\tkzSetUpLine[line width=2pt]
\tkzDrawLines[add=12.5 and 2.5](A,h)
\tkzDrawLines[-Latex, add=0 and 1.5](A,a)
\tkzDrawSegment[blue,dash pattern = on 6pt off 3pt on 1pt off 3pt on 1pt off 3pt](O,B)
\tkzDrawSegment[green,dashed](O,A)
\tkzDrawSegment[-Latex,green](O,A')
\tkzDrawSegment[-Latex,blue](O,B')
\tkzDrawCircles[line width=2pt](O,P)
\tkzDrawArc[line width=2pt,delta=0,orange](N,D)(C)
\tkzDrawArc[line width=2pt,delta=0,orange](M,F)(E)
\tkzDrawArc[line width=2pt,delta=0,orange](J,E)(D)
\tkzDrawArc[line width=2pt,delta=0,orange](O,C)(F)
\tkzMarkRightAngles(a,A,B h,B,O)
\tkzLabelPoint[below left](O){$O$}
\tkzLabelPoint[above](A){$A$}
\tkzLabelPoint[above right](B){$B$}
\tkzText[below right=20pt](M){$K$}
\tkzText[above=60pt](O){$\mathcal{S}^{n-1}$}
\tkzText[below left=60pt](J){$\rho(K,v)$}
\tkzText[shift={(-20pt, 70pt)}](A){$H_{u,t}$}
\tkzText[shift={(80pt, 20pt)}](A){$u=\alpha^{*}_{K}(v)$}
\tkzText[shift={(60pt, -60pt)}](B){$u$}
\tkzText[shift={(10pt, -10pt)}](O){$u$}
\tkzText[shift={(0pt, 10pt)}](O){$v$}
\draw [red, decorate,decoration={brace, amplitude=80pt},
xshift=4pt,yshift=0pt]
(B) -- (O) node[black,midway,shift={(80pt, -80pt)}]
{$h(K,u)$};
\end{tikzpicture}
\end{document}结果为：
使用需要lualatex的tkz-elements包(v5.13c)在处理椭圆切线等问题会方便很多：
\documentclass[border=5pt]{standalone}
\usepackage[mini]{tkz-euclide}
\usepackage{tkz-elements}
\directlua{
init_elements()
z.O = point(0.15, 0.6)
C.OW = circle(through(z.O, 0.5))
z.W = C.OW.through
z.M = point(-0.25, -1)
z.N = point(1,4)
z.C = tkz.midpoint(z.M, z.N)
L.MN = line(z.M, z.N)
z.T = L.MN:orthogonal_at(z.C, {length = 1.5})
CO.EL = conic(EL_points(z.C, z.N, z.T))
PA.curve = CO.EL:points(0, 1, 100)
z.P = point(4,-1)
L.tan, _ = CO.EL:tangent_from(z.P)
z.Q = L.tan.pb
L.normal = L.tan:ortho_from(z.Q)
z.dir = L.normal:report(-1.5, z.Q)
z.H = line(z.P, z.Q):projection(z.O)
z.v, _ = intersection(line(z.O,z.Q), C.OW)
z.u, _ = intersection(line(z.O,z.H), C.OW)
}
\begin{document}
\begin{tikzpicture}
\tkzGetNodes
\tkzDrawCoordinates[smooth,thick](PA.curve) \tkzDrawCircle[black,thick](O,W)
\tkzDrawSegments[thick,-latex](Q,dir)
\tkzDrawSegments[thick,add=0 and 0.5](P,Q)
\tkzDrawSegments[thick,dashed,green](O,Q)
\tkzDrawSegments[thick,green,-latex](O,v)
\tkzDrawSegments[thick,dash dot,blue](O,H)
\tkzDrawSegments[thick,blue,-latex](O,u)
\tkzMarkRightAngle[size=.25,thick](O,H,Q)
\tkzMarkRightAngle[size=.25,thick](P,Q,dir)
\draw [magenta,decorate,decoration={brace,mirror,amplitude=10pt}] (O) -- (H) node[midway,below right,xshift=-.15cm,yshift=-.25cm,black] {$h(K,u)$};
% labels left is your time~
\end{tikzpicture}
\end{document}
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您好！这个图能不能请您写一下代码呢？
其实用
tkz-elements宏包会更方便，基于lua内置了一个conic类