Let AOB be the cable of uniformly loaded suspension bridge. Let AL and BM be the longest wires of length 30 m each. Let OC be the shortest wire of length 6 m and LM be the roadway. Now AL = BM = 30 m, OC = 6 m and LM = 100 m ∴ LC = CM = $\frac{1}{2}$ LM = 50 m Let O be the vertex and axis of the parabola be y-axis. So, the equation of parabola in standard form is $x^2$ = 4ay Coordinates of point B are (50, 24) Since point B lies on the parabola $x^2$ = 4ay ∴ $(50)^2$ = 4a × 24 ⇒ a = $\frac{2500}{4 \times 24}$ = $\frac{625}{24}$ So, equation of parabola is $x^2$ = $\frac{4 \times 625}{24}$ y ⇒ $x^2$ = $\frac{625}{6}$ y Let length of the supporting wire PQ at a distance of 18 m be h. ∴ OR = 18 m and PR = PQ – QR = h – 6. Coordinates of point P are (18, h – 6) Since the point P lies on parabola $x^2$ = $\frac{625}{6}$ y ∴ $(18)^2$ = $\frac{625}{6}$ (h - 6) ⇒ 324 × 6 = 625h - 3750 ⇒ 625h = 1944 + 3750 ⇒ h = $\frac{5694}{625}$ = 9.11 m approx.