<s>[INST] Asymptotes of $(x(t),y(t)) = \bigg(\frac{1+t^2}{2+t^3}, \frac{t}{2+t^3}\bigg)$, collinear points, ... Consider the curve:
\begin{equation}
(x(t),y(t)) = \bigg(\frac{1+t^2}{2+t^3}, \frac{t}{2+t^3}\bigg)
\end{equation}
Question 1: What are his asymptotes?
Answer: In projective space: $[(2+t^3,1+t^2,t)]$. The intersection with the line at infinity ($x_0=0$), is $t= -(2)^{\tfrac{1}{3}}$. But how to check that this gives an asymptote and not a parabolic direction?
Question 2: Give a necessary condition such that $t_1,t_2,t_3$ gives collinear points.
Answer: I put it in a determinant, but then it's difficult to find a condition. Are there better ways to do this?
Question 3: Show that the curve has a 'turning point' (point with multiplicity two where the two tangent lines are the same)
Answer: I have no idea.
 [/INST] To  the third question: Entering into M2
R=QQ[x,y]
S=QQ[t]
KS=frac S
f=map(KS,R,{(1+t^2)/(2+t^3),t/(2+t^3)})
ker f

we see $2x^3-6x y^2+5y^3-x^2-2x y+2y^2+y=0$ is the implicit equation. The partial derivatives 
$6x^2-6y^2-2x-2y,-12xy+15y^2-2x+4y+1$
have a common zero at $(x,y)=(2/3,1/3)$ which is on the curve. This is your singular point ($t=1$). Moving this point to the origin; the tangent cone is the lowest degree term of $2(x+2/3)^3-6(x+2/3) (y+1/3)^2+5(y+1/3)^3-(x+2/3)^2-2(x+2/3) (y+1/3)+2(y+1/3)^2+(y+1/3)=5 y^3-6 x y^2+3y^2-6x y+2x^3+3x^2$ which is $3(y^2-2xy+x^2)=3(y-x)^2$.
</s>