I solved it generally
Let w = half the distance between the poles
Let h = the vertical distance from the lowest point of the rope to the top
Let l = the length of the rope
Then
w=ln((l+2h)/(l-2h))*(l^2-4*h^2)/(8*h)
Working:
Equation of catenary:
y=a*cosh(x/a) + c
Gradient of catenary:
dy/dx=sinh(x/a)
Length of curve is equal to the sum of the lengths of all the infinitely small lines making it:
l=SUM(sqrt(dy*dy+dx*dx)) `Pythagoras
dy=sinh(x/a)*dx
l=SUM(sqrt(sinh(x/a)^2*dx^2+dx^2))
l=SUM(sqrt((sinh(x/a)^2+1)*dx^2))
l=SUM(sqrt(sinh(x/a)^2+1)*dx)
SUM(f(x)*dx) = $ f(x) dx `($ = integral sign)
l=$ (sinh(x/a)^2+1)^0.5 dx
(Integrate this using the substitution u=x/a)
Therefore, the length of a catenary between x_1 and x_2 is given by:
l=a*sinh(x_2/a) - a*sinh(x_1/a)
We want the length from -w to w, so:
l=a*sinh(w/a) - a*sinh(-w/a)
l=a*sinh(w/a) + a*sinh(w/a)
l/2=a*sinh(w/a)
Height at lowest point of catenary is at x=0:
y=a*cosh(x/a) + c
y=a*cosh(0) + c
y=a + c
Height at pole is at x=w:
y=a*cosh(w/a) + c
Difference in height:
h=(a*cosh(w/a) + c) - (a + c)
h=a*cosh(w/a)-a
We now have simultaneous equations:
l/2=a*sinh(w/a)
h=a*cosh(w/a)-a
Let b=w/a:
l/2=a*sinh(b)
h=a*cosh(b)-a
Add equations:
l/2+h=a*(sinh(b)+cosh(b)-1)
l/2+h=a*(e^b-1) `sinh(x)+cosh(x) = e^x
Subtract equations:
l/2-h=a*(sinh(b)-cosh(b)+1)
l/2-h=a*(1-e^-b) `sinh(x)-cosh(x) = -e^-x
a=(l/2-h)/(1-e^-b)
l/2+h=a*(e^b-1)
Substitute in a:
l/2+h=(l/2-h)*(e^b-1)/(1-e^-b)
l/2+h=(l/2-h)*e^b
e^b=(l/2+h)/(l/2-h)
Hooray, solved for b!
b=ln((l+2h)/(l-2h))
Substitute in b:
a=(l/2-h)/(1-e^-b)
a=(l/2-h)/(1-e^-ln((l+2h)/(l-2h)))
a=(l/2-h)/(1-(l-2h)/(l+2h))
Hooray, solved for a!
a=(l^2-4*h^2)/(8*h)
From definition:
b=w/a
w=b*a
Hooray, solved for w!
w=ln((l+2h)/(l-2h))*(l^2-4*h^2)/(8*h)
Also, given the conditions of the question, that is not actually a catenary curve. As you said yourself, a catenary curve is defined as:
y=a*cosh(x/a)
For the rope to hang as you described, 'a' must be equal to zero, which it obviously cannot be!
[b]
