I need to find the distance from the point provided in the hyperboloid model with a vector $x$ where $\langle x,x\rangle=-1$ to the hyperplane $H_e$ with a normal vector $e$, where $\langle e,e\rangle=1$.
I have some kind of solution it is here:
$\rho$ is the distance, and $r_e$ is defined as $r_e: x \mapsto x - 2\langle e,x\rangle e$.
$\rho(x,H_e) = \frac12\rho(x,r_e(x))$
$\cosh(\rho(x,r_e(x))) = -\langle x,r_e(x)\rangle = -\langle x, x - 2\langle e,x\rangle e\rangle = 1 + 2\langle e,x\rangle^2$
$\cosh(2\times \frac12\rho(x,r_e(x))) = 1 + 2 \times \sinh^2(\frac12\rho(x,r_e(x)))$
Hence, $\sinh(\rho(x,H_e)) = \lvert\langle e,x\rangle\rvert$.
Is it correct? Is there a way to simplify it?
The way I understand your question (and I hope my edit to it was correct), you intend $r_e$ to denote a reflection in $H_e$. So this is my main concern, whether that operation is really a reflection with respect to the hyperbolic metric. It is easy to see that every point on $H_e$ itself will remain fixed under this operation, since $\langle e,x\rangle=0$ for these. In addition to this, you need the property that ideal points (i.e. light-like vectors) will be preserved:
$$\langle y,y\rangle=0\;\implies\;\langle r_e(y),r_e(y)\rangle=0$$
When I first posted this answer, I proposed a counterexample. But I must have gotten my math wrong, in more than one place. Now I am convinced that this is actually true, due to the following consideration:
$$ \langle r_e(x),r_e(x)\rangle = \bigl\langle x-2\langle e,x\rangle e,x-2\langle e,x\rangle e\bigr\rangle = \langle x,x\rangle -4\langle e,x\rangle^2 +4\langle e,x\rangle^2\langle e,e\rangle = \langle x,x\rangle $$
So a point on the hyperboloid gets correctly mapped to a point on the hyperboloid. Therefore I trust that $r_e$ is indeed a reflection. After that, everything looks watertight to me.
As for simplification: neither the formula nor its deduction seem to leave a lot of room for simplifications. There might be slightly shorter ways to write down the deduction, depending on what prior knowledge you may assume.