TITLE: Geometry behind calling the pair hyperbolic in bilinear space
QUESTION [7 upvotes]: The name hyperbola is originally comes from an object in geometry, but several other objects in mathematics wear partly the name hyperbola or hyperbolic; for example, hyperbolic Mobius transformation. There are some geometric reasons for calling them hyperbolic.
In the study of symplectic spaces (i.e. a vector space $V$ with a biliear form $(\cdot,\cdot)$ with $(u,u)=0$), a pair of vectors $\{u,v\}$ is called hyperbolic if $(u,v)=1$. 
However, the books (or notes or sites) which give this definition of hyperbolic pair do not mention the geometric reason behind hyperbolic. 
I tried to search in the books with title Linear algebra and geometry (lot of books are there with this title) the reason for it, but I didn't succeed! 
[I didn't even find it in some books on Linear algebra written by famous geometers - Sahafarevich, Dieudonne,...]
Can one explain a little the reason behind calling the pair $\{u,v\}$ with $(u,v)=1$ hyperbolic?

REPLY [5 votes]: There are three types of second order curves in $\Bbb R^2$:

ellips (in particular, circle), e.g. $x^2+y^2=1$,
hyperbola, e.g. $x^2-y^2=1$,
parabola, e.g. $y=x^2$.

In general, all three types can be represented as level curves $f(x,y)=\text{const}$ of a quadratic function 
$$
f(x,y)=\begin{bmatrix}x\\y\end{bmatrix}^TH\begin{bmatrix}x\\y\end{bmatrix}+c^T\begin{bmatrix}x\\y\end{bmatrix}.
$$
One gets (see also discriminant of a conic section)

ellips iff $H$ has two eigenvalues of the same sign,
hyperbola iff $H$ has two eigenvalues of opposite signs,
parabola iff $H$ has one zero eigenvalue and $\operatorname{rank}([H\  c])>\operatorname{rank}(H)$.

Traditionally, mathematical objects that are described by a quadratic form inherit the same names depending on the signature of the corresponding matrix $H$ (elliptic/hyperbolic/parabolic PDE, elliptic/hyperbolic geometry etc)
In your case, a hyperbolic pair are two linear independent vectors $u$, $v$ with those particular properties. If the "distance" is defined by the associated quadratic form $q(w)=B(w,w)=w^THw$ then the unit "circle" $q(xu+yv)=1$ becomes $2xy=1$, which is a hyperbola. Note that the matrix $H$ must be indefinite, in particular, in the plane $uv$ the quadratic form with the matrix $\begin{bmatrix}0 &1\\1 & 0\end{bmatrix}$ having one positive and one negative eigenvalues, i.e. the subspace $uv$ with the metric $q$ is hyperbolic.