What is the proof of
$$\gamma_{\mathbf{x} \rightarrow \mathbf{u}}(t) = \cosh(t)\mathbf{x} + \sinh(t)\mathbf{u}$$
being the formula for a geodesic on the hyperboloid model of hyperbolic space? I haven't been able to find a source that makes it clear. It seems that always the sinh and cosh come out of nowhere so I am just trying to understand how they follow from the definition of the hyperboloid.
Let $B$ be the Minkowski bilinear form and assume that $x$ and $u$ satisfy \begin{align*} B(x,x) &= -1\\ B(u,u) &= 1\\ B(x,u) &= B(u,x) = 0 \end{align*} A curve that lies in intersection of the hyperboloid with the $2$-plane spanned by $x$ and $u$ is of the form $$ c(t) = a(t)x + b(t)u, $$ where \begin{align*} -1 &= B(c(t),c(t)). \end{align*} Observe that the velocity of $c$ is always space-like. The curve has unit speed with respect to the Minkowski metric, if \begin{align*} 1 &= B(c'(t), c'(t)). \end{align*} I'll leave as an exercise the proof that the unique solution to the equations above that also satisfies $c(0)=x$ and $c'(0)=u$ is \begin{align*} c(t) = x\cosh t + u\sinh t. \end{align*}