I got stuck at exercise 5.26 on Fulton's Algebraic curves, 2008, which reads as follows:
Well, following the clues, it suffices to show it when $P=[0:1:0]$ (by a linear change of coordinates). Since every line that passes through $P$, except for the line at infinity, is of the form $L_\lambda$ for some $\lambda$, then we must focus on these lines. Since $P=[0:1:0]$, we can write $F=\sum_{i=m_P(F)}^nA_i(X,Z)Y^{n-i}$ with $A_{r}\neq 0$, because $m_P(F)=m_{(0,0)}(F(X,1,Z))$.
So, in order to find $L_{\lambda}\cap F$, we must solve the equation $F\left (\lambda,t,1\right )=0$ for $t$. This is because we do not have to worry about any point on infinity since the only point at infinity in $L_{\lambda}$ is $P$, and we are not interested in $P$ but in the other points of $L_{\lambda}\cap F$.
So, maybe we should show that $G_{\lambda}(t)$ and $\frac{\partial}{\partial t}G_{\lambda}\left (t\right )=\sum_{i=r}^n\frac{\partial A_i}{\partial y}(\lambda, t)$ do not have any roots in common if $A_r\left (\lambda,1\right )\neq 0$ and $\deg G_{\lambda}=n-r$; and since the characteristic of $k$ is zero, then we would be done.
I am not succeeding in proving it. Maybe the irreducibility of $F$ implies easily that this is the case, but I sincerely cannot see it.
How would you solve the exercise?
The irreducibility of form $F$ holds even when dehomogenized. As a result, the two polynomials $G$ and the lower $t$-degree $G_t$ (suppressing the $\lambda$ subscript) have no common factors in $k(\lambda)[t]$ due to Gauss lemma. Euclidean division gives factors $A,B \in k(\lambda)[t]$ such that $AG + BG_t = 1$. $A,B$ are not defined at a finite number of $\lambda$ values due to zero denominators. Discarding these values, if $G, G_t$ have a common zero, we get $0=1$.