An amazing property of the Catenary I discovered that if we want an arc of catenary  in the interval $[a,b]$ we solve
$$\int_a^b \sqrt{\cosh '(x)^2+1} \, dx=\int_a^b \cosh x \, dx$$
which means that the "result" of the length is equal to the result of the area in the same interval, though in different units.
So I asked myself if there are any other curve with the same property.
I set
$$y=\sqrt{y'^2+1}\to y^2=y'^2+1; y(0)=1$$
then
$$y'=\sqrt{y^2-1}\to dx=\frac{dy}{\sqrt{y^2-1}}\to x=\cosh^{-1} \,y$$
hence, arbitrary constant is zero,
$y=\cosh x$
But I am not sure how to deal with the other solution $y'=-\sqrt{y^2-1}$ even if Mathematica   gives the same result $y=\cosh x$
I'd like someone checking  this proof, you know: I am not a pro, I am just an (almost) retired high school teacher :)
Update 9/1/2020. Now I am officially retired :)
 A: EDIT1:
I understood  your question in this way:
How is it that the area under a Catenary is proportional to the arc length? i.e., how is
$$ c=\dfrac{A}{L}$$
valid for  some constant of proportionality $c$?
At first about the sign in  front of radical sign in DE
Let us at the outset consider  very familiar similar situations:
If two  DEs  are given as
$$ y'= + \sqrt {1-y^2},\; y'= - \sqrt {1-y^2} $$
we have in either case by squaring
$$ y^{'2} = (1-y^2) $$
Differentiate
$$ 2 y' y^{''}= -2 y y',\to y^{''}+y =0 $$ which is the differential equation of a sine curve.
With BC $ x=0,y=1,y'=0  \to y= \cos x $ in either case
Similarly if two  DEs  are given as
$$ y'= + \sqrt {y^2-1},\; y'= - \sqrt {y^2-1} $$
we have in either case
$$y^{'2}= (y^2-1)$$
Differentiating
$$ 2 y' y^{''}= 2 y y',\to y^{''}-y =0 $$ which is the differential equation of a Catenary. With BC $ x=0,y=1,y'=0  \to y= \cosh x $ in either case.
However,  if you do not wish to square thereby losing its sign but wish to directly integrate the two BCs, the following:
$$ y'= + \sqrt {1-y^2},\; y'= - \sqrt {1-y^2} $$
we get
$$ \sin^{-1}y= x +c_1,  \sin^{-1}y=- x-c_2 $$
$$y= \sin (x+c_1),y= -\sin (x+c_2)$$
For (an even ) a symmetric solution $ x=0, y=1 $
we have respectively
$$c_1=\pi/2, c_2= 3 \pi/2$$
both yielding the same solution
$$ y = \cos x $$
When we have here our actual case
$$ y'= + \sqrt {1+y^2},\; y'= - \sqrt {1+y^2} $$
we get
$$ \cosh^{-1}y= x +c_1,  \cosh^{-1}y=- x-c_2 $$
$$y= \cosh (x+c_1),y= \cosh (x+c_2)$$
For even symmetric solution $ x=0, y=1 $
we have respectively
$$c_1= c_2= 0 $$
yielding  both the same solution
$$ y = \cosh x $$
So we can say in conclusion that in front of any (square root) radical sign we have $\pm$ and both signs are equally applicable for first order DE. It is only by a convention that we put in a positive sign implying the unsaid negative.  They result in the same differential equation and hence also the same integrand for given boundary conditions in this particular case.
Geometrically a negative or positive sign of derivative relates to different slopes of the curve in different portions of the curve.
Next to answer what I considered to be your main  question let us set up its DE which uniquely defines the curve.
To  get a physical/geometrical idea a length dimension quantity $c$ is introduced as the quotient of covered area $A$ to the length of its curved "roof".
$$c=\dfrac {\int y \; dx}{\int\sqrt{1+y'^2}dx}$$
Using Quotient Rule differentiate to simplify
$$c=\dfrac{ y} {\sqrt{1+y'^2}}= \to y' = \dfrac{\sqrt{y^2-c^2}}{c} $$
which is the differential equation of the unique curve being sought.
Integrating with boundary condition $ y(0)=c ,y'(0)=0,$  one obtains the equation of the only curve that satisfies the required property.
$$ \dfrac{y}{c}= \cosh\dfrac{x}{c}$$
which is recognized as a catenary as stated. And in association this property is recognized as well... that $c$ is the constant of proportionality which is the minimum distance of the  catenary to the x-axis.
$$ c=\dfrac{A}{L}$$
as also shown here graphically.

A: From
$$\frac{y'}{\sqrt{y^2-1}}=\pm1$$ you draw
$$\text{arcosh}(y)=c\pm x$$
and
$$y=\cosh(c\pm x).$$
With the initial condition $y(0)=1$,
$$y=\cosh(\pm x)$$ which is $$y=\cosh(x).$$
