All catenarys are given by the hyperbolic cosine function
f(x) = acosh(x/a) = 1/2 (ex/a + e-x/a)
for part a) the lowest point of a catenary is always the vertex, which is at (0,a) in an orthogonal coordinate system. Hence, in this case its at (0,1) since a = 1 (given)
b) If the endpoints are...