Def: A hyperbolic curve is an algebraic curve obtained by removing $r$ points from a smooth, proper curve of genus $g,$ where $g$ and $r$ are nonnegative integers such that $2g−2+r > 0.$
How does this relate to a hyperbola, which is an algebraic curve? I don't understand the "removing points" part. Why would you remove points?
Thanks for clearing up my confusion.
This is actually a nice question. The reason that such algebraic curves are called hyperbolic is that when you look at the corresponding complex algebraic curve regarded as a Riemann surface $X$, the surface $X$ admits a (unique in the given conformal class) complete hyperbolic metric, of constant curvature $-1$ (necessarily of finite area) compatible with the conformal structure. In contrast, when $2g-2+r\le 0$, $X$ does not admit a complete hyperbolic metric. However, it admits either a complete flat metric or, when $g=0, r=0$, it admits a spherical metric. Now, why hyperbolic surfaces are called hyperbolic is a separate question. One reason maybe is because of the hyperboloid model of the hyperbolic plane. However, it appears that the terminology hyperbolic plane was first introduced by Felix Klein in 1871, before the hyperboloid model was known. From Wikipedia:
Confusingly though, complex algebraic curves/Riemann surfaces admitting a compatible complete Euclidean metric are not called Euclidean, they are called elliptic curves for some historic reasons, going back to the early 19th century (study of elliptic integrals).
As for the question "why would you remove points," the answer varies: People do this for different reasons. For instance, maybe instead of considering a complex projective curve, you want to study an affine complex algebraic curve $X$, for instance $X$ which is the solution set of an equation $$ P(z,w)=0, $$ where $(z,w)\in {\mathbb C}^2$ and $P$ is a polynomial. Then $X$ is obtained by removing some points from the corresponding complex projective curve. Or, as a complex analyst, maybe you have a compact Riemann surface $Y$ and a meromorphic function $f$ on $Y$. You then want to identify the domain of $f$ which is the surface $X$ obtained from $Y$ by removing the poles of $f$, i.e. removing a finite subset of $Y$. Or, maybe, as an algebraic geometer, you want to study the moduli space of stable curves in the given algebraic variety $Z$. Then the finite subset of a projective curve $Y$ will appear as the set $F$ of marked points whose images under maps $Y\to Z$ belongs to, say, given subvarieties in $Z$. The hyperbolicity condition for $X=Y-F$ is equivalent to the condition that automorphism group $Aut(Y,F)\cong Aut(X)$ is finite.