I am given the curve $G = (x_1-x_2)^3 - x_0 x_1^2$ and want to compute all of its regular points whose tangents go through $[0:0:1]$.
Firts, the line $L = ax_0 + bx_1 + cx_2$ has to go through $[0:0:1]$.$ \implies c = 0 $ and hence $L = ax_0 + bx_1$.
Since are looking for the intersection of a line with the curve G in a regular point, the intersection multiplicity will be $1$. I can parametrize $G$ as $[t_0, t_1] \mapsto [(t_0-t_1)^3:t_0^3:t_0^2t_1]$. If I substitute this in $L$ I get: $a(t_0-t_1)^3 + bt_0^3$ and then I should somehow be able to get the tangents and the multiplicities.
But I am not sure if I am going the right way, and if so, how to proceed.
$(x_1-x_2)^3-x_0x_1^2=0$ has dual $27 X_0 X_1^2+54X_0X_1X_2+27X_0X_2^2-4X_2^2=0$ and $(x_0:x_1:x_2)=(0:0:1)$ corresponds to $X_2=0$ via $x_0X_0+x_1X_2+x_2X_2=0.$ Intersecting we get $\langle 27 X_0 X_1^2+54X_0X_1X_2+27X_0X_2^2-4X_2^2, X_2\rangle=\langle X_2, X_0X_1^2\rangle$ i.e. $(X_0:X_1:X_2)=(0:1:0)$ corresponding to $x_1=0$ and $(X_0:X_1:X_2)=(1:0:0)$ (doubly) corresponding to $x_0=0.$ The tangents through $(x_0:x_1:x_2)=(0:0:1)$ are therefore $x_0=0$ and $x_1=0.$ But only the latter to a regular point of the curve.