It seems to me that the answer should be yes, but my naive attempts to come up with an example have failed.


Just to clarify, by finite hyperbolic geometry I mean a finite set of points and lines such that


1. every pair of points determines determines a unique line
2. every line contains at least two points
3. there are at least two distinct lines
4. for every line L and every point p not on the line, there are at least two distinct lines which are incident with p which do not intersect L


It is easy enough to show that in finite euclidean geometries (i.e. affine planes) all lines have the same number of points, and in finite elliptic geometries in which every line contains at least $3$ points (i.e. projective planes) all lines have the same number of points. 


It is also easy enough to come up with an example of a finite hyperbolic geometry in which not every line has the same number of points (but some lines have exactly two points). e.g. Let the points be $\{1,2,3,4,5,6\}$ and let the lines be the set $\{1,2,3\}$, together with all of the $2$-element subsets except for $\{1,2\}$, $\{1,3\}$, $\{2,3\}$.



Does there exist a finite hyperbolic geometry in which every line contains at least 3 points, but not every line contains the same number of points?
Take a 2-$(v,4,1)$ design on $v$ points and delete one block, along with the four points on it. In the original system each point is on exactly $(v-1)/3$ blocks, so if we assume $v\ge25$ the geometry we get by deleting one block is hyperbolic, all blocks have at least three points, and there are blocks of size three and size four.


