Hyperbolic rotation of spacetime and Lorentz transformation My question is:
What is the motivation behind deriving Lorentz transformation using hyperbolic functions? Is it because the formulation in such way offers a handy mathematical tool? Or is  there something else in special relativity that needs such derivation? From what I know, it is nothing but a treatment of the event co-ordinates in way that resembles a normal co-ordinate transformation in xyz axes. 
Any clarification would be helpful.  
 A: It is. The quantity $ds^2=-cdt^2 +dx^2+dy^2+dz^2$ must be constant for all observers, in the same way as the radius $r^2=x^2+y^2+z^2$ must be the same for any classical observer (Euclidean transformations).
The opposite sign of the time coordinate makes the circular rotation invariance become hyperbolic invariance.
A: You can easily write down Lorentz transformations in terms of $\beta$ (relative speed in $c$ units) and $\gamma$ (the relativistic coefficient), but the composition is not trivial (velocity don't compose linearly anymore in relativity, as they do instead in Galilean transformations).
Instead of this one can define the rapidity $w$ such that $\cosh w = \gamma$ and you find again the linear composition rule.
https://en.m.wikipedia.org/wiki/Rapidity
Try to multiply Lorentz matrices in the two forms (in terms of $\beta(v)$- $\gamma(v)$ and in terms of $w$ and find yourself).
P.S.: For simplicity do this try in 2D space-time (1 dimensional physical space and the time, such as $t-x$): it is completely general the same, because composing such a transformation with the rotations you can obtain all the others.
A: Because the speed of light must be the same in every inertial frame of reference, one can write:
$$c=\frac{dx}{dt}=\frac{dx’}{dt’}$$
So we can define the space-time interval:
$$ds^2=c^2dt^2-dx^2= c^2dt’^2-dx’^2=ds’^2$$
Which is invariant in every inertial frame of reference. In the (+ - - -) metric signature convention.
$$ds^2=
\begin{cases}
< 0  & space-like \\[2ex]
= 0  & light-like \\[2ex]
> 0  & time-like
\end{cases}
$$
In the (- + + +) metric signature convention.
$$ds^2=
\begin{cases}
< 0  & time-like \\[2ex]
= 0  & light-like \\[2ex]
> 0  & space-like
\end{cases}
$$
In the (- + + +) metric signature convention, one can write $ dx^2-c^2{dt}^2$ as $ dx^2+(icdt)^2$. This looks like $dx^2+dy^2$ which is basically the length of some vector in Cartesian coordinate system. Now imagine that you rotate the $(x,y)$ coordinate system by some angle $\theta$. The rotation does not change the length of the vector. So the main idea is that the space-time interval is invariant under Lorentz transformations in the same way as the length of that vector is invariant under rotations of the coordinate system.
The dot product of any four-vectors is Lorentz invariant just like the dot product of any vector with itself in Euclidean space is the same regardless if you rotate the coordinate system.
A: Since the other answers so far mostly tackle the question of why the hyperbolic rotations do properly represent the Lorentz transformations, I'd like to write a few lines about why they might give a better intuition than the usual representation in terms of the velocity.
The analogy to usual rotations gives a very nice intuition to effects like time dilatation and length contraction. If we consider any vector in two-dimensional euclidean space, its components will behave in the following way under rotations: One component will decrease, the other will increase. This is exactly what happens with lengths and time intervals in SR. 
So time dilatation and length contraction, effects which seem mysterious to beginners, get a very intuitively accessible analogy to corresponding effects known from rotations which we observe in our daily life.
