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hyperbolic rotation
Let $\mathbb{E}$ be the Euclidean plane equipped with the Cartesian coordinate system. Recall that given a circle $C$ centered at the origin $O$, one can define an “ordinary” rotation $R$ to be a linear transformation that takes any point on $C$ to another point on $C$. In other words, $R(C)\subseteq C$.
Similarly, given a rectangular hyperbola (the counterpart of a circle) $H$ centered at the origin, we define a hyperbolic rotation (with respect to $H$) as a linear transformation $T$ (on $\mathbb{E}$) such that $T(H)\subseteq H$.
Since a hyperbolic rotation is defined as a linear transformation, let us see what it looks like in matrix form. We start with the simple case when a rectangular hyperbola $H$ has the form $xy=r$, where $r$ is a nonnegative real number.
Suppose $T$ denotes a hyperbolic rotation such that $T(H)\subseteq H$. Set
$\begin{pmatrix}x^{{\prime}}\\ y^{{\prime}}\end{pmatrix}=\begin{pmatrix}a&b\\ c&d\end{pmatrix}\begin{pmatrix}x\\ y\end{pmatrix}$
where $\begin{pmatrix}a&b\\ c&d\end{pmatrix}$ is the matrix representation of $T$, and $xy=x^{{\prime}}y^{{\prime}}=r$. Solving for $a,b,c,d$ and we get $ad=1$ and $b=c=0$. In other words, with respect to rectangular hyperbolas of the form $xy=r$, the matrix representation of a hyperbolic rotation looks like
$\begin{pmatrix}a&0\\ 0&a^{{1}}\end{pmatrix}$
Since the matrix is nonsingular, we see that in fact $T(H)=H$.
Now that we know the matrix form of a hyperbolic rotation when the rectangular hyperbolas have the form $xy=r$, it is not hard to solve the general case. Since the two asymptotes of any rectangular hyperbola $H$ are perpendicular, by an appropriate change of bases (ordinary rotation), $H$ can be transformed into a rectangular hyperbola $H^{{\prime}}$ whose asymptotes are the $x$ and $y$ axes, so that $H^{{\prime}}$ has the algebraic form $xy=r$. As a result, the matrix representation of a hyperbolic rotation $T$ with respect to $H$ has the form
$P\begin{pmatrix}a&0\\ 0&a^{{1}}\end{pmatrix}P^{{1}}$
for some $0\neq a\in\mathbb{R}$ and some orthogonal matrix $P$. In other words, $T$ is diagonalizable with $a$ and $a^{{1}}$ as eigenvalues ($T$ is nonsingular as a result).
Below are some simple properties:

Let $P$ be the pencil of all rectangular hyperbolas centered at $O$. For each $H\in P$, let $[H]$ be the subset of $P$ containing all hyperbolas whose asymptotes are same as the asymptotes for $H$. If a hyperbolic rotation $T$ fixing $H$, then $T(H^{{\prime}})=H^{{\prime}}$ for any $H^{{\prime}}\in[H]$.

$[\cdot]$ defined above partitions $P$ into disjoint subsets. Call each of these subset a subpencil. Let $A$ be a subpencil of $P$. Call $T$ fixes $A$ if $T$ fixes any element of $A$. Let $A\neq B$ be subpencils of $P$. Then $T$ fixes $A$ iff $T$ does not fix $B$.

Let $A,B$ be subpencils of $P$. Let $T,S$ be hyperbolic rotations such that $T$ fixes $A$ and $S$ fixes $B$. Then $T\circ S$ is a hyperbolic rotation iff $A=B$.

In other words, the set of all hyperbolic rotations fixing a subpencil is closed under composition. In fact, it is a group.

Let $T$ be a hyperbolic rotation fixing the hyperbola $xy=r$. Then $T$ fixes its branches (connected components) iff $T$ has positive eigenvalues.

Suppose $T$ fixes the unit hyperbola $H$. Let $P,Q\in H$. Then $T$ fixes the (measure of) hyperbolic angle between $P$ and $Q$. In other words, if $\alpha$ is the measure of the hyperbolic angle between $P$ and $Q$ and, by abuse of notation, let $T(\alpha)$ be the measure of the hyperbolic angle between $T(P)$ and $T(Q)$. Then $\alpha=T(\alpha)$.
The definition of a hyperbolic rotation can be generalized into an arbitrary twodimensional vector space: it is any diagonalizable linear transformation with a pair of eigenvalues $a,b$ such that $ab=1$.
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Special relativity
It is forth pointing out that the transformations discussed
here play an important role in physics, where they often go
under the name of "boosts". Namely, in the Special Theory of
Relativity, they describe how one transforms a frame of
reference into the frame of reference of an observer moving at
uniform velocity.
Re: Special relativity
You mean 'worth'? Raymond. (my English, you know)