I'm studying William Fulton's "Algebraic Curves," and I'm currently studying projective plane curves. One of the problems in the section asks us to find the tangents to $xy^4+yz^4+xz^4$ at its multiple points. However, nowhere in the section does he define what the tangent is.
The only thing he comments on is, "We can define a line L to be tangent to a curve F at P if $I(P, F \bigcap L)>m_P(F).$" Where $m_P(F)$ denotes the multiplicity of $F$ at $P$, and $I(P,F\bigcap L)$ is the intersection number of $F$ and $L$ and $P$. How am I supposed to use this to find the tangent lines to a given curve?
I know how to do this with an affine plane-curve, and my intuition was to find a tangent line in the affine case at the level curve $Z=1$, and then find the plane which contains that line, and define it as the tangent line to $xy^4+yz^4+xz^4$, but based off of examples I've been trying to piece together online, it seems like this is the incorrect definition.
It can be shown (by taking the gradient of $f=xy^4+yz^4+xz^4$) that the multiple points of $V(f)$ are of the form $[x:0:0]$ where $0 \ne x \in K$ ($K$ an algebraically closed field). Let $P=[1:0:0] \in \mathbb{P}^2.$ Note that there are more multiple points than this, but the tangents are easy to find once this first case is established (it is just cumbersome to type). Then, $f(1,y,z)=y^4+yz^4+z^4$. We will consider this as an affine variety, taking its projective closure at the end. Since $P=[1:0:0]$, in $\mathbb{A}^2$ we will consider the point $(0,0)$, and to find the tangent lines we will consider the leading for of $f(1,x,y)$, which is $y^4+z^4$. The leading form factors as follows: $$y^4+z^4=(y^2+\omega z^2)(y^2- \omega z^2)$$ where $\omega$ denotes $(-1)^{1/2} \in K$. Then, the above simplifies further to $$(y-\omega^{1/2}z)(y+\omega^{1/2}z)(y-(-\omega)^{1/2}z)(y+(-\omega)^{1/2}z). $$ Thus, each of those components are the tangents to $f(1,y,z)$. Each component is already homogenous, and so homogenizing the components (taking the projective closure) has no effect. It then follows that each component in the factorization of $y^4+z^4$ is tangent to $f$ at $[1:0:0]$.