Boundary slope of an incompressible surface in a knot exterior

Assume full AC. Let $K\subset S^3$ be a smooth oriented knot with chosen compact exterior $E_K$ and boundary torus $T_K=\mathrm{\partial }{E}_K$. A simple closed curve on a surface is essential if it is not null-homotopic there. All surfaces considered here are compact. A surface $S\subset E_K$ is properly embedded if $S\cap \mathrm{\partial }{E}_K=\mathrm{\partial }S$ and its interior lies in the interior of $E_K$. It is incompressible if it has no embedded disk with interior disjoint from $S$ and essential boundary in $S$; it is boundary-incompressible if there is no embedded disk $D\subset E_K$ with $\mathrm{\partial }D=\alpha \cup \beta$, where $\alpha ,\beta$ are arcs meeting exactly at their common endpoints, $D\cap S=\alpha$, $D\cap \mathrm{\partial }{E}_K=\beta$, and $\alpha$ does not cobound a disk in $S$ with an arc of $\mathrm{\partial }S$. A boundary slope of $K$ is the unoriented isotopy class of an essential simple closed curve on $T_K$ that occurs as a boundary component of a connected, properly embedded, orientable, incompressible, boundary-incompressible surface. Using the positive meridian $\mu$ and preferred longitude $\lambda$ as a homology basis, a primitive class $a[\lambda ]+b[\mu ]$ has slope $b\mathrm{/}a\in \mathbb{Q}\cup \{\mathrm{\infty }\}$, with $a=0$ interpreted as $\mathrm{\infty }$. A boundary slope is strict if it has a connected realizing orientable, incompressible, boundary-incompressible surface which is not a fiber of any locally trivial fibration $E_K\to S^1$; a fiber is the inverse image of a point under that bundle projection.

Hatcher, Boundary Slopes of Incompressible Surfaces

Baldwin and Sivek, §4.1