Show that for the catenary y=a cosh x/a, the ordinate
Show that for the catenary $y=a\cosh\dfrac{x}{a}$, the ordinate at any point is the geometric mean of the constant $a$ and the length of the normal at that point, i.e.\ that $y_0^2=a\cdot N$.
Explanation of Method
The identity cosh²−sinh²=1 turns 1+(y′)² into a perfect square over a².
Step 1: The length of the normal
For $y=f(x)$, the length of the normal at $(x_0,y_0)$ is $N=y_0\sqrt{1+y'(x_0)^2}$.
Step 2: Differentiate and use the hyperbolic identity
\[ y' = \sinh\frac{x}{a}. \tag{1} \]
At $(x_0,y_0)$ with $y_0=a\cosh\dfrac{x_0}{a}$ and $y_0'=\sinh\dfrac{x_0}{a}$, the identity $\cosh^2t-\sinh^2t=1$ gives
\[ 1+(y_0')^2 = 1+\sinh^2\frac{x_0}{a} = \cosh^2\frac{x_0}{a} = \frac{y_0^2}{a^2}. \tag{2} \]
Step 3: Compute the normal
\[ N = y_0\sqrt{1+(y_0')^2} = y_0\cdot\sqrt{\frac{y_0^2}{a^2}} = y_0\cdot\frac{y_0}{a} = \frac{y_0^2}{a}. \tag{3} \] so $a\cdot N=y_0^2$, i.e.\ $y_0=\sqrt{a\cdot N}$: the ordinate is the geometric mean of $a$ and the normal length. $\blacksquare$
Step 4: Verification
\[a=2,\ x_0=1.3\text{: } y_0\approx2.19329,\ N=y_0^2/a\approx2.40529,\ \sqrt{aN}\approx2.19329=y_0\ \checkmark\]
Result
Final answer:
\[\boxed{\,y_0^{2} = a\cdot N, \qquad\text{i.e.}\qquad y_0 = \sqrt{a\cdot N}\,}\]
Remark
Remark. The subtangent $T=y_0/y_0'=a\coth(x_0/a)$ does not satisfy $T^2=a\cdot N$ in general (it, too, depends on $x_0$ in a way that does not cancel against $N$); the identity that holds unconditionally, and that (2)–(3) prove directly, is $y_0^2=a\cdot N$ — the geometric-mean statement is about the ordinate $y_0$, not the subtangent.