Geometry behind calling the pair hyperbolic in bilinear space 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?
 A: 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.
