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This question is related to this question I asked previously. In the answer, the user gives finds the asymptotes of the algebraic curves $C$ and $D$ defined by $$f = -X^4+X^3Y+Y, \, g = -X^3 + X^2Y+Y$$ respectively, and apparently, none of the curves have parabolic branches. The definition he used though is not the one I am using, and when I tried to do it the way I know I had some problems. First, the definitions I'm using are

Definition: Let $C \subset \mathbb{C}^2$ be an algebraic curve and $\overline{C} \subset \mathbb{P}^2(\mathbb{C})$ its projective completion, and $L_{\infty}$ the line at infinity. Then a line $L$ is an asymptote of $C$ at an infinity point $P_{\infty}$ if its projective completion $\overline{L}$ is tangent to $\overline{C}$ at $P_{\infty}$. If there is some point at infinity $P_{\infty}$ of $C$ such that $L_{\infty}$ is tangent to $P_{\infty}$ at $\overline{C}$, then $C$ posses parabolic branches associated at $P_{\infty}$.

So first of all, the points at infinity are $P_1 = (0:1:1)$ and $P_2=(0:0:1)$. Now, $P_1$ is regular, and we can compute the tangent line of $\overline{C}$ at $P_1$ easily: $$X_0F_0(0:1:1)+X_1F_1(0:1:1)+X_2F_2(0:1:1)=X_2-X_1=0$$ where $F$ is the homogeneous polynomial associated to $f$. Since $X_2 - X_1=0$ is not the line at infinity, $X_0 = 0$,that gives us the asymptote $X = Y$. The computations for the curve defined by $g$ are the same, and again, there is an asymptote $X=Y$.

The problem appears for the second point, $P_2$, which is singular, and therefore the tangent curve at $P_2$ of $\overline{C}$ doesn't make sense the way I found it for $P_1$. I believe I need to find the equation of the tangent cone as suggested in this question and comments, but $P_2$ is a point at infinity, and I only know how to find the equation of the tangent cone when the point is on the affine plane (e.g, $(1:0:0) \sim (0,0)$ or $(1:2:3) \sim (2,3)$). That raises my first question

How do I find the tangent line to a point at infinity that is singular?

I'm not really sure if this is the way to go, or even if the question above makes sense, but I need to find an asymptote of equation $X+1= 0$ of $C$, and I don't know how using only the definition I am using. Thank you.

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  • $\begingroup$ If you look at $-x^4+(x^3+z^3)y$ and set y=1, you get the tangent cone (the lowest degree terms) $(x^3+z^3)=(x+z)(x^2-xz+z^2)$ of which we only see $(x+z)$ over the reals. $\endgroup$ – Jan-Magnus Økland Aug 22 '17 at 8:18
  • $\begingroup$ @Jan-MagnusØkland Thank you for the comment. I get that i need to set $y=1$ because the point is $(0:0:1)$ and that $x+z$ ends in $x+1$, but why is it that we only see $x+z$ over the reals? Shouldn't the other factor play some role too? $\endgroup$ – user313212 Aug 22 '17 at 11:27
  • $\begingroup$ Over ${\Bbb C}$ the other two roots to $x^3=-1$ also give asymptotes, yes. They are just hard to plot. $\endgroup$ – Jan-Magnus Økland Aug 22 '17 at 11:31
  • $\begingroup$ @Jan-MagnusØkland Thanks again! Just out of curiosity, is there anything interesting in those extra two asymptotes even though I can't plot them on the real plane? $\endgroup$ – user313212 Aug 22 '17 at 12:11
  • $\begingroup$ I really don't know. $\endgroup$ – Jan-Magnus Økland Aug 22 '17 at 12:16

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