I believe that the focal locus is the same thing as the conjugate locus in Riemannian geometry. Given a codimension 1 submanifold $S \subset \mathbb{R}^n$, you have a naturalthe exponential map $S \times \mathbb{R} \rightarrow \mathbb{R}^n$ is given by $(x, t) \mapsto t\gamma(x)$$(x, t) \mapsto x + t\gamma(x)$, where $\gamma$ is the Gauss map. The cut locus is the closure of all points in $\mathbb{R}^n$ that have more than one pre-image. The focal points is the closure of all points where the map is not a diffeomorphism.
Given any point where the curvature $\kappa$ is nonzero on a smooth curve in the plane, there is a corresponding point on the focal locus at distance $1/|\kappa|$ on the side of the curve that is inwardly curved.
I believe that the focal locus is always contained in the cut locus.
ADDED: (Corrected definition of map above)...If you differentiate the exponential map (the one I define above), then since the differential of the Gauss map is the second fundamental form, call it $A$, then you see that if a point $y$ lies in the focal locus, there exists $x \in S$, $t \in \mathbb{R}$, and a nonzero $v \in \mathbb{R}^n$ such that $v + tAv = 0$. I neglected to say above that the focal locus corresponds to the closest point on either side of $S$ along the geodesic normal to $x$ where this equation holds. Therefore, there is a focal point on the half line where $t > 0$ only if there is a negative principal curvature and the focal point is at distance $-1/\kappa$, where $\kappa$ is the negative principal curvature with largest magnitude. There is a focal point on the other half line only if there is at a positive principal curvature, and it is at distance $1/\kappa$, where $\kappa$ is the largest positive principal curvature.