(ii) (a) - y² 12 = a² => (a + h)²−(LAB/2)² = a² for Is = LAB/2=10m -2a+ √4a²-4(-12) That is h² + 2ah−(LAB/2)² = 0 => h = · at point B, we have y = H = a + h = 10 +4.14 = 14.14 m, and a=10 m. = 4.14m 2 (b) as, y = acosh(), and a=10 m and y = 14.14 m D x 2 = acosh¹ (#) = 10cosh 1 (14.14) = 8.88 m, D = 17.6 m A LAB H a D y B s(x, y) h X Figure 5.1 The graph of a catenary in the Cartesian coordinates is shown in Figure 5.1. The two poles, A and B, have the same height to the x-axis, and the same distance to the y-axis. For a given point on the catenary, s(x, y), we have, y = acosh(). (1) It is also known that the length of the catenary segment between C (the lowest point of the catenary) and point s, i.e., Is, is given by, y² - 12 = a². (2) If the total length of the catenary between A and B, i.e., LAB, is 20 m and a is 10 m, determine (i) the height of the pole, i.e. H, as shown in the figure. (ii) the distance between A and B, i.e., D, as shown in the figure.