I am given a curve $F: 2XY^2-3Z^3 = 0$ and a line $L:2X+6Y-9Z=0$ both in the projective plane. I am asked to calculate all the intersection points $F \cap L$, their multiplicities and tangents to $F$ at the regular intersection points.
Using the method I have been shown I have parameterised $L$ with two points $P = \left[3,-1,0\right]$ and $Q=\left[9,0,2\right]$ giving;
$$L: \left[S,T\right] \mapsto \left[3S+9T, -S, 2T\right] \:.$$
Now I calculate
$$F(3S+9T, -S, 2T): 6(S-T)(S+2T)^2 = 0$$
Now I see that I have a simple intersection at $[S,T] = [1,1]$ which gives me the point $P_1 = [12,-1,2]$. From this I believe I can calculate the tangent at $P_1$ using the partial derivatives $F_X, F_Y$ and $F_Z$ evaluated at $P_1$ giving;
$$L_1 : 2X + 48Y -36Z=0 \: .$$
My problem is how do I evaluate the other point corresponding to when $(S-2T)^2 = 0$. I assume I can say that there is an intersection at $[S,T] = [1,-\frac{1}{2}]$ corresponding to the point $P_2 = [\frac{3}{2},1,1]$ however what is the multiplicity of this point? Also would these two points be all the intersection points? Many many thanks!