The line x = t (with t > 0) meets the curves y = a sinh

The line $x = t$ (with $t > 0$) meets the curves $y = a\sinh\dfrac{x}{a}$ and $y = a\cosh\dfrac{x}{a}$ at points $M_1$ and $M_2$ respectively, and meets the $x$-axis at point $M$. Show that the length of segment $MM_1$ equals the arc length of $y = a\cosh\dfrac{x}{a}$ from $T(0,a)$ to $M_2$.

We compute both quantities and compare.

Step 1: Identify the points

$M=(t,0)$, $M_1=\left(t,\,a\sinh\frac{t}{a}\right)$, $M_2=\left(t,\,a\cosh\frac{t}{a}\right)$, and $T=(0,a)$ since $a\cosh0=a$.

Step 2: Length of $MM_1$

\[|MM_1|=a\sinh\frac{t}{a}\qquad(\text{positive for }t>0)\]

Step 3: Arc length of the catenary from $T$ to $M_2$

For $y=a\cosh\frac{x}{a}$, $y'=\sinh\frac{x}{a}$, so using $1+\sinh^2u=\cosh^2u$:

\[\sqrt{1+(y')^2}=\sqrt{1+\sinh^2\frac{x}{a}}=\cosh\frac{x}{a}\]

\[L=\int_0^t\cosh\frac{x}{a}\,dx = a\left[\sinh\frac{x}{a}\right]_0^t = a\sinh\frac{t}{a}\]

Step 4: Compare

Both quantities equal $a\sinh\frac{t}{a}$, so

\[\boxed{|MM_1|=L_{T\to M_2}=a\sinh\frac{t}{a}}\]

Step 5: Interpretation

The catenary is the shape of a uniform chain hanging under gravity. This identity says the arc length measured from the lowest point of the chain out to any point equals the vertical height, at that same horizontal position, of the associated hyperbolic-sine curve — a classical property connecting $\cosh$ and $\sinh$.