Tracing on a plane the concurrent tangent lines of a surface

Question

Let $X$ be a smooth cubic algebraic surface in $\mathbb P^3(K)$, where $\mathrm{char} K = 0$. Is it true that if I take a point $P\in X$ and a plane $E$ general enough then the insterection points (I'd like to think of this as a "trace") of the lines with $E$ that cross $P$ and tangent to $X$ will be a smooth curve? I guess if $X$ is of degree $>3$ this is not necessarily true (there may be self-intersection of this curve). If so, what is the argument for this? I'd like to say that this is due to Bertini's theorem but I do not see this immediately. Also, is there a way to generalize this for $X$ of higher degree?

Answer 1

The morphism $$ \mathrm{Bl}_P(X) \to \mathbb{P}^2 $$ induced by the linear projection from $P$ is a flat double covering. Its ramification divisor $R \subset \mathrm{Bl}_P(X)$ is isomorphic to its branch divisor $B \subset \mathbb{P}^2$, and is equal to the fixed locus of the Galois involution of $\mathrm{Bl}_P(X)$. Therefore, it is smooth.