How does one apply the formula for $(ds)^2$ on the hyperbolic half-plane? We commonly say that the metric on the upper half-plane model of the hyperbolic plane is given by
$$(ds)^2=\frac{(dx)^2+(dy)^2}{y^2}.$$
(There are a lot of articles that start this way and I've been guilty of writing it myself.)
The $ds$
means the length of an arc in the half-plane.
The $dx$
means the differential of this arc with respect to horizontal movement.
The $dy$
means the differential of this arc with respect to vertical movement.
What does the denominator mean?
The $y$-coordinate is changing as we move along a curve so what does a variable mean here?
I definitely waited way too long in my life to ask this question. Not being a differential geometer I always accepted this formula as saying that distance in the half-plane changes according to the reciprocal of the height. Honestly I probably could figure this out but it would be nice to have a better explanation than I can give, online, for the sake of others studying the hyperbolic plane.
If a person tries to understand how this gives the hyperbolic metric via Wikipedia,
for instance,
s/he finds a metric defined as a distance function, which this is not.
If the person tries to chase that down, s/he ends up reading about metric tensors, the first fundamental form, Gaussian curvature, etc., which should not be necessary to understand the formula ... and regardless there are no examples there where a formula for $(ds)^2$ has a variable in it without a differential.
With that motivation in mind, let's try to keep the answer as simple as possible. To make the question more concrete, here are some basic specific questions (which you can answer or not answer as you see fit to explain the general idea).


*

*Use the formula to show that a curve that approaches a point on the real line has infinite length.

*Use the formula to show that the hyperbolic geodesics are the vertical half-lines and the half-circles perpendicular to the real line.

*Use the formula to find the length of some curve, let's say $\big\{(t+1,t^2+1)\mid t\in[0,1]\big\}$.

*Use the formula to derive the actual half-plane metric,
which is $$d_{\mathcal{H}^2}\big((x,y),(x',y')\big)=\mathrm{arcosh}\Big(1+\frac{(x'-x)^2+(y'-y)^2}{2yy'}\Big).$$

 A: It does more than just give you lengths: it also gives you angles.  This is really shorthand for $\frac{1}{y^2} (dx \otimes dx +dy\otimes dy)$.  This is a Riemannian metric on the upper half plane:  at any point $(x,y)$ you can feed it two tangent vectors $v_1$ and $v_2$ and it will spit out a ``dot product'' of these two vectors.  In this case, it is just the usual dot product, but scaled by the reciprocal of the square of the height.
Given limited time I will just answer your third question.  The length of a tiny line segment in this metric should be the square root of the dot product of the vector with itself.  We use the usual definition of arc-length of a curve, substituting the new notion of dot product:
$$
\int_0^1 \sqrt{\frac{1}{(t^2+1)^2} \left[\left(\frac{d}{dt}(1+t) \right)^2 + \left(\frac{d}{dt}(t^2+1) \right)^2\right]} dt
$$
A: Here's the short answer:

The formula $ds^2 = \frac{dx^2 + dy^2}{y}$ defines a norm in the tangent plane to $\Bbb{H}^2$ at $(x,y)$.

A bit of elaboration.
By polarization we can produce a dot product in each tangent plane. The length of a curve is defined by integrating the lengths of its tangent vector field. From the dot product we can produce (locally minimizing) geodesics in the method of Riemannian geometry and it is an exercise to show that in $\Bbb{H}^2$, by extending any locally minimizing geodesic one must produce a circle or line perpendicular to the real axis.
A: Interesing question , I tried to make a complete proof to proof that in the distance function $ds$ the dominator has a functor $y$ there 
Proofsketch that y is a factor in de denominator of the distance function in the Poincare halfplane model.
This proof is made out of 3 parts:
1) a proof that Poincare halfplane model is a model of (non metric) incidence hyperbolic geometry
2) a proof that in hyperbolic geometry similar triangles are congruent
3) a proof that in the Poincare halfplane model similar triangles can have different euclidean arc lengths
1) a proof that Poincare halfplane model is a model of (non metric) incidence hyperbolic geometry.
This proof is simple you only need to proof:
- Any pair of points is on one and only one hyperbolic line
- there is a point $P$ not on a line $l$ and through $P$ there are two lines $p_1$ and $p_2$ not intersecting $l$
2) a proof that in hyperbolic geometry similar triangles are congruent
for this we need a method to compare angles, for the  Poincare halfplane model we can just take the angles between the tangents at the vertex point 
3) a proof that in the Poincare halfplane model similar triangles can have different euclidean arc lengths
For 2 and 3 we can use the following construction:
1 draw a half circle (hyperbolic line) $c_1$ centered on the X-axis
2 on $c_1$ choose two points $A$ and $B$ (keep $A$ and $B$ close to one end of the halfcircle)
3 through $A$ draw the line $l$ perpendicular to the X-axis
4 the intersection of $l$ and the X-axis is the point $\Omega$
The ray $l$ , $\Omega A$ is also a hyperbolic line
5 draw a half circle (hyperbolic line) $c_2$ centered on  point $\Omega$ going through $B$ 
$c_2$ is also a hyperbolic line
6 the intersection of $l$ and $c_2$ is the point $C$
We have now constructed a hyperbolic triangle $\triangle ABC$
To proof that in hyperbolic geometry similar triangles are congruent we can do:
7 Choose a point $E$ on $c_1$ so that $B$ is between $A$ and $E$ 
8 draw a half circle (hyperbolic line) $c_3$ centered on  point $\Omega$ going through $E$ 
$c_3$ is also a hyperbolic line
9 the intersection of $l$ and $c_3$ is the point $F$
We have now constructed a second hyperbolic triangle $\triangle AEF$ 
$\triangle AEF$  is bigger than $\triangle ABC$  and
$\angle BAC = \angle EAF$, $\angle ACB = \angle AFC$
but $\angle ABC \not= \angle AEF$
proving that triangles  that are not are congruent are not similar.
10 draw the euclidean line $e_1$ through $\Omega$  and $B$
(this is not an hyperbolic ) 
11 on $e_1$ choose a point  $L$  so that $B$ is between $\Omega$  and $G$ (so that $\Omega B $ is about the same length as $BG$ )
10 draw the euclidean line $e_2$ through $L$  and parallel to the  euclidean segment $AB$
11 the intersection of $l$ and $e_2$ is the point $K$
12 draw a half circle (hyperbolic line) $c_4$ centered on  point $\Omega$ going through $L$ 
$c_4$ is also a hyperbolic line
13 the intersection of $l$ and $c_4$ is the point $M$
14 draw a half circle (hyperbolic line) centered on  a point on the x-axis going through the points $K$ and $L $
We have now constructed a hyperbolic triangle $\triangle KLM$ 
but 
$\angle BAC = \angle LKM$ , $\angle ACB = \angle LMK$
and $\angle ABC = \angle KLM$
but this means that the triangles are similar  and thus congruent and thus the sides have the same lenghts.
thus $d_h(AB) =d_h(KL) $, $d_h(BC) =d_h(LM) $, $d_h(AC) =d_h(KM) $
especially $d_h(BC) =d_h(LM) $ is importand because
the euclidean angle $\angle B\Omega C = \angle L\Omega M $
