Soit Δ la droite d'équation x=2a, C un cercle de centre (-a,0) et de rayon a, et H un point courant de la droite Δ. On définit le pôint P comme l'intersection de la droite (OH) avec le cercle C privé du point O. On détermine ensuite M comme le milieu du segment [HP]. Quand H décrit Δ, M décrit la cissoïde droite.
verbatimtex %&latex \documentclass{article} \usepackage[garamond]{mathdesign} \usepackage{amsmath} \usepackage[utf8]{inputenc} \usepackage[frenchb]{babel} \begin{document} etex u:=1cm; a:=3u; path cercle, droite, cissoide; pair M,O; M:=(-a,0); O:=(0,0); droite:=(2a,-8u)--(2a,8u); cercle:= fullcircle scaled 2a shifted M; for i:=0 upto 400: beginfig(i); path drtour,demid; pair M,P,Hb,H; drawarrow (-8u,0)--(8u,0); drawarrow (0,-8u)--(0,8u); H:=(2a,i/10*u-20u); Hb:=(-xpart H, -ypart H); demid:= 0.01[O,Hb]--2[O,Hb]; P:=demid intersectionpoint cercle; M:=.5[H,P]; if i=0: cissoide:=M; else: cissoide:=cissoide--M; fi; draw cercle withpen pencircle scaled 0.8pt withcolor blue; draw droite withpen pencircle scaled 0.8pt withcolor blue; draw H--P dashed evenly withpen pencircle scaled 0.8pt withcolor green; draw cissoide withpen pencircle scaled 1pt withcolor red; dotlabel.urt(btex $M$ etex,M); dotlabel.rt(btex $H$ etex,H); dotlabel.llft(btex $P$ etex,P); dotlabel.lrt(btex $O$ etex, O); label.bot(btex $x$ etex, (7.6u,0)); label.rt(btex $y$ etex, (0,7.6u)); label.bot(btex $2a$ etex, (-a,0)); label.bot(btex $2a$ etex, (a,0)); label.urt(btex $\mathcal{C}$ etex, (-0.5a,a)); label.rt(btex $\Delta$ etex, (2a,-7.5u)); label.bot(btex \textit{Cissoïde droite} : \(\left\{\begin{array}{l} x=\dfrac{2at^2}{1+t^2}\\[2mm] y=\dfrac{2at^3}{1+t^2} \end{array}\right.\) etex,(-4u,6.3u)); clip currentpicture to (-8u,-8u)--(-8u,8u)--(8u,8u)--(8u,-8u)--cycle; endfig; endfor; end.