Assume a heavy chain (constant mass per unit length) takes the shape of a plane curve $\mathcal C$ after being suspended by its two ends from the same height. Let $s$ be its arc length starting from its lowest point, let $T$ be the tension in the chain, let $\varphi$ be the angle between the tangent line and the horizontal, and let $\lambda$, $\mu$ be two non-zero constants.
I'm trying to show that $T\cos \varphi = \lambda $ and $T\sin \varphi = \mu s$.
Just as a gut check, if we plug in $s=0$ where the tangent line is horizontal so $\varphi = 0$, we find that $T=\lambda\neq 0$, but to me the tension in the chain at its lowest point or where $s=0$ should be zero since there is no chain below it to hold up.
Without solving the problem for me here, where am I going wrong?
Draw a free body diagram of forces. Consider the portion of the chain between the lowest point and another point P above.
Tension T at P above is inclined, so has two components. One vertical component $Q=T\sin\varphi$ is balanced by the chain weight and the horizontal component of chain weight $T\cos \varphi$ is counteracted or balanced by an always constant non-zero horizontal force at P = H say.
The differential equation of catenary is $$ \tan \varphi= \frac{Q}{H}=\frac{\mu s}{H}~$$ if $\mu$ is the chain weight per unit arc length. Hope you take it from there.