with(linalg): F:=t*x^2-y^3;

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H:=det(hessian(F,[x,y,t]));

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solve({H=0, F=0}, {x,y,t});

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X:=(1,0,0); # e' l'unico flesso, O=(0,0,1) e' un punto doppio, anzi e' l'unica cuspideCpolare:=(xx,yy,zz) -> multiply(<x|y|t>, multiply((subs({x=xx,y=yy,z=zz},hessian(F,[x,y,t])),<x,y,t>)));

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grad(F,[x,y,t]);

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Cpolare(X);

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S:=x*(2*x+2)+2*x; # e' la componente della conica polare che non passa per X

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solve({S=0, F=0}, {x,y,t});

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#Quindi abbiamo ottenuto solo un punto Q=(-2,1,1/4) non singolare, (O e' l'unico punto singolare di F)#Quindi le rette tangenti, passanti per X sono solo t=0, e XQ.clear;

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#Ripetiamo il discorso per:F:=x*y*t-x^3-y^3; # O=(0,0,1) e' un nodo ed e' l'unico punto singolare

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H:=det(hessian(F,[x,y,t]));

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solve({H=0, F=0, t=1}, {x,y,t});solve({H=0, F=0, t=0}, {x,y,t});

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solve(Z^2+1-Z=0, Z);

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z1:=1/2+1/2*I*3^(1/2); z2:=1/2-1/2*I*3^(1/2);

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P1:=(-1,1,0); P2:=(z1,1,0); P2:=(z2,1,0); # sono i tre flessi

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CP1:=Cpolare(P1);

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solve({CP1=0, F=0,t=0}, {x,y,t});

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# (-1,1,0) e' P1, e quindi lo scartiamo solve({CP1=0, F=0,t=1}, {x,y,t});

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solve(-4*_Z^2-1+28*_Z^4=0,_Z); # (0,0,1) e' l'unico punto singolare e quindi lo scartiamo. # In definitiva, abbiamo 2 punti semplici. Quindi le rette tangenti passanti per P sono 3.

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CP2:=Cpolare(P2); solve({CP1=0, F=0,t=0}, {x,y,t}); solve({CP2=0, F=0,t=1}, {x,y,t});

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