Asymptotes of $(x(t),y(t)) = \bigg(\frac{1+t^2}{2+t^3}, \frac{t}{2+t^3}\bigg)$, collinear points, ...

Question

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.

Answer 1

\begin{cases} x(t) = \frac{1+t^2}{2+t^3}\\ y(t) = \frac{t}{2+t^3} \end{cases}

ASYMPTOTE :

$x$ and $y$ tend to infinity when $\begin{cases}(2+t^3)\to 0 \\ t\to -2^{1/3} \end{cases}$

Let $t=-2^{1/3}+\epsilon \quad$ the series expansion leads to : $$\begin{cases} x=\frac{1}{6}(2^{1/3}+2)\frac{1}{\epsilon} -\frac{1}{6}(2^{2/3}-1) +O(\epsilon) \\ y=-\frac{1}{6}2^{2/3}\frac{1}{\epsilon}+\frac{1}{18}\epsilon +O(\epsilon^2) \end{cases}$$ $\frac{y}{x}=-\frac{2^{1/3}}{1+2^{2/3}} -\frac{2^{2/3}-1}{(1+2^{2/3})^2}\epsilon +O(\epsilon^2) \quad\to\quad \frac{y}{x}=-\frac{2^{1/3}}{1+2^{2/3}} -\frac{2^{2/3}-1}{(1+2^{2/3})^2}\frac{1}{6}(2^{1/3}+2)\frac{1}{x} +O(\epsilon^2)$ $y=-\frac{2^{1/3}}{1+2^{2/3}}x -\frac{2^{2/3}-1}{(1+2^{2/3})^2}\frac{1}{6}(2^{1/3}+2)+O(\epsilon)$ $$y=-\frac{2^{1/3}}{1+2^{2/3}}x -\frac{2-2^{1/3}}{6(1+2^{2/3})} +O(\epsilon)$$ Thus, the equation of the asymptote is : $$y=-\frac{2^{1/3}}{1+2^{2/3}}x -\frac{2-2^{1/3}}{6(1+2^{2/3})}\quad\to\quad y=ax+b \quad \begin{cases} a\simeq -0.486945 \\b\simeq -0.047672 \end{cases}$$

TURNIG POINT :

The study of the derivatives $\begin{cases}x'(t)=\frac{(4-3t-t^3)t}{(2+t^3)^2} \\y'(t)=\frac{2(1-t^3)}{(2+t^2)^2}\end{cases}$ shows that both $x$ and $y$ reaches a local maximum at $(t=1 \:,\:x=\frac{2}{3} \:,\: y=\frac{1}{3})$ which is characteristic of a turning point.

NECESSARY CONDITION,

such that $t_1 ,t_2 ,t_3 $ gives collinear points :

The three points must be on a straight line $y=\alpha x + \beta$ . Hense $t_1 ,t_2 ,t_3 $ must be roots of the equation : $$\frac{t}{2+t^3} = \alpha \frac{1+t^2}{2+t^3} +\beta$$ $\alpha(1+t^2)+\beta (2+t^3)-t=0$

$$t^3+\frac{\alpha}{\beta} t^2 -\frac{1}{\beta}t+\frac{\alpha}{\beta}+2=0$$

$\begin{cases} t_1+t_2+t_3=-\frac{\alpha}{\beta} \\ t_1t_2+t_2t_3+t_3t_1=-\frac{1}{\beta}\\ t_1t_2t_3=-(\frac{\alpha}{\beta}+2) \end{cases} \quad\to\quad \begin{cases}\alpha=\frac{t_1+t_2+t_3}{t_1t_2+t_2t_3+t_3t_1} \\ \beta =-\frac{1}{t_1t_2+t_2t_3+t_3t_1}\\ t_1t_2t_3=t_1+t_2+t_3-2 \end{cases}$

The necessary condition is : $\quad t_1+t_2+t_3-t_1t_2t_3=2$

FIGURE :

Answer 2

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 $3(y^2-2xy+x^2)=3(y-x)^2$.