Let $X\subset \mathbb{P}^3$ be an irreducible quartic surface, defined over an algebraically closed field $k$. Suppose that $X$ is rational (i.e. birational to $\mathbb{P}^2$). Is is true that $X$ has one of the following type of singularities?

a) a point $q\in X$ of multiplicity $4$ (hence $X$ is a cone over a rational quartic curve).

b) a point $q\in X$ of multiplicity $3$ (hence the projection away from $q$ gives a birational map $X\dashrightarrow \mathbb{P}^2$).

c) a curve of singular points.

It seems to me that these are the only possible cases, as otherwise the canonical divisor would be trivial. For instance, if $X$ has only isolated double points, the canonical divisor is trivial. But maybe one could have infinitely near curves of singularities that make the canonical divisor non-effective? Is it possible to get a rational surface?

It seems to me that this should be quite classical. I looked over the internet for "rational quartic surfaces" and found a lot of examples and some articles of classifications, but I did not find the answer to the above question. Thanks for your help.