Exercice 3On d\351fnit la param\351trisation, puis on construit le support partiel et sir la totalit\351 des param\350tresrestart;x:=t->3*t^2;y:=t->2*t^3;NiM+SSJ4RzYiZio2I0kidEdGJUYlNiRJKW9wZXJhdG9yR0YlSSZhcnJvd0dGJUYlLCQqJDkkIiIjIiIkRiVGJUYlNiM+SSJ5RzYiZio2I0kidEdGJUYlNiRJKW9wZXJhdG9yR0YlSSZhcnJvd0dGJUYlLCQqJDkkIiIkIiIjRiVGJUYltotal:=[x(t),y(t),t=-100..100]: plot(total);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 le vecteur vitesse VV:=unapply([D(x)(t),D(y)(t)],t);NiM+SSJWRzYiZio2I0kidEdGJUYlNiRJKW9wZXJhdG9yR0YlSSZhcnJvd0dGJUYlNyQsJDkkIiInLCQqJEYuIiIjRi9GJUYlRiU=M\352me pour M(0) la tangente en M(t) est Y=t(X-x(t))+y(t)Cherchons les points d'intersection de Tt avec Gammaequ:=y(tau)=t*(x(tau)-x(t))+y(t);solve(equ,tau);NiM+SSRlcXVHNiIvLCQqJEkidEdGJSIiJCMhIiIiIiVGJw==NiMsJEkidEc2IiMhIiIiIiM=tau:=-t/2;NiM+SSR0YXVHNiIsJEkidEdGJSMhIiIiIiM=On veut que la tangente en M(tau) soit orthogonale \340 la tangente en M(t), i.e vu que les pentes sont tau et teq:=tau*t=-1;solve(eq,t);t:='t':NiM+SSNlcUc2Ii8sJCokSSJ0R0YlIiIjIyEiIkYqRiw=NiQsJCokIiIjIyIiIkYlISIiRiQ=2 solutions:logique car on a une sym\351trie / OxOn \351crit la tangente en t0=2^1/2 et la tangente en tau_0=-2^1/2/2 (et le sym\351trique)t_0:=sqrt(2):xT1:=x(t_0)+t:yT1:=y(t_0)+t_0*t:EqT1:=[xT1,yT1,t=-infinity..infinity]:tau_0:=-sqrt(2)/2:xN1:=x(tau_0)+t:yN1:=y(tau_0)+tau_0*t:EqN1:=[xN1,yN1,t=-infinity..infinity]:t_0:=-sqrt(2):xT1:=x(t_0)+t:yT1:=y(t_0)+t_0*t:EqT2:=[xT2,yT2,t=-infinity..infinity]:tau_0:=sqrt(2)/2:xN2:=t:yN2:=tau_0*(t-x(tau_0))+y(tau_0):EqN2:=[xN2,yN2,t=-infinity..infinity]:total:=[x(t),y(t),t=-infinity..infinity]: 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