Let $C = V(f)$ and $D=V(g)$, where $$f = Y^2 - X^2 + X^3$$ $$g = Y^2 - X^3 + X^4$$ and let be $\overline{C}$ and $\overline{D}$ be their respective projective completions. I'd like to know how to answer the following questions
How many vertical tangent lines do $C$ and $D$ have? What are the multiplicity of intersection of $D$ and each branch of $C$ at $(0,0)$ and $(1,0)$. How many branches have $C$ and $D$ at the infinity?
I've already prove that both $\overline{C}$ and $\overline{D}$ are irreducible and that $C$ is parametrizable. For the question about the number of vertical lines of $C$ and $D$, I thought about tracing the polar curve through $(0:0:1)$ on each curve and then intersecting it with the curve itself. Doing that, I get that $$P_C = V\left(\frac{\partial F}{\partial X_2}\right) = 2X_0X_1$$ where $F = X_0^2X_2^2 - X_0X_1^2 + X_1^3$ is the homogenization of $f$. Then, intersecting with $V(F)$, I get that if $X_0 = 0$ then $X_1^4 = 0$ so I find the point $(0:0:1)$ with multiplicity $4$, and if $X_2 = 0$, then I find the points $(1:0:0)$ with multiplicity $2$ and $(1:1:0)$ with multiplicity $1$. Does that mean that the number of vertical tangent lines is the sum of the multiplicities of points I have found? That is, does $C$ have $7$ vertical tangent lines?
Doing the same with $D$ instead, I get the same points with different multiplicities. In this case I get $(0:0:1)$ with multiplicity $4$, $(1:0:0)$ with multiplicity $3$ and $(1:1:0)$ with multiplicity $2$.
About the second question, I thought about parametrizing $C$ and then see what happens at the points $(1:0:0)$ and $(1:1:0)$. So I let $T_0 = X_1$, $T_1 = X_2$ and when substituting in $F$ I get $$X_0 = \frac{T_0^3}{T_0^2 - T_1^2}$$ so a polynomial parametrization of $C$ would be given by $$X_0 = T_0^2,\ X_1 = T_0(T_0^2 - T_1^2), \ X_2 = T_1(T_0^2 - T_1^2)$$ Now, substituting the parametrization on $G$, the homogenization of $g$ ($G = X_0^2X_1^2 - X_0X_1^3 + X_1^4$) I get: $$G(X_0, X_1, X_2) = T_0^4T_1^4(T_0-T_1)^2$$ Now I need to find what values of the parameters $(T_0:T_1)$ I get the points $(1:0:0)$ and $(1:1:0)$. For the origin, I get $$(T_0:T_1)=(1:1), \; (T_0:T_1) = (1:-1)$$ both with multiplicity $2$, and for $(1:1:0)$ I get $$(T_0:T_1) = (1:0)$$ with multiplicity $4$.
Finally, for the last question, I first found the point at infinity, that for both curves is $(0:0:1)$. When finding the values for $T_0$ and $T_1$ in the parametrization of $C$, I find $(T_0:T_1) = (0:1)$, which means that I only have one branch with multiplicity $4$ at $(0:0:1)$, right? The problem is when trying to find the branches at $D$ since it seems I can't fin a polynomial parametrization by following the same steps as before.
If anyone could give me a hint about how to proceed or point me the mistakes I made, I'd be grateful. Thank you!
EDIT: I forgot to mention, but the base field is $\mathbb{C}$, that is, both $C$ and $D$ are complex algebraic curves.