Horizontal reaction of cable, derive H formula

Data: cable of span $L$ carrying a UDL $w$ per unit horizontal length; the supports are at heights $h_1$ and $h_2$ above the lowest point of the cable. Let $x_1$ and $x_2$ be the horizontal distances from the lowest point to the supports, so $x_1 + x_2 = L$.

 support 1 o                       o support 2
            \                     /  ^
          h1 \                   /   | h2
             |\___.___lowest___.'     
             |<x1->|<---x2---->|

Equation of the cable

Take the origin at the lowest point $O$, where the tension is horizontal and equal to $H$. For a point $P$ at horizontal distance $x$ from $O$, take moments about $P$ of the forces on the portion $OP$ of the cable:

$$H\,y = \frac{w\,x^2}{2} \Rightarrow y = \frac{w\,x^2}{2H}$$

Apply to the two supports

$$h_1 = \frac{w\,x_1^2}{2H} \Rightarrow x_1 = \sqrt{\frac{2H\,h_1}{w}}, \qquad h_2 = \frac{w\,x_2^2}{2H} \Rightarrow x_2 = \sqrt{\frac{2H\,h_2}{w}}$$

Since $x_1 + x_2 = L$:

$$\sqrt{\frac{2H}{w}}\,\left(\sqrt{h_1}+\sqrt{h_2}\right) = L$$

Squaring both sides:

$$\frac{2H}{w}\left(\sqrt{h_1}+\sqrt{h_2}\right)^2 = L^2$$ $$\boxed{H = \frac{w\,L^2}{2\left(\sqrt{h_1}+\sqrt{h_2}\right)^2}}$$

This is the horizontal reaction at each end, since the cable carries no horizontal external load. Check: for $h_1 = h_2 = h$, $H = \dfrac{wL^2}{2\times 4h} = \dfrac{wL^2}{8h}$, the usual result for supports at the same level.