Let AB be the parabolic arch having O as the vertex and OY as the axis. The parabola is of the form $x^2$ = 4ay. Now, CD = 5 m ⇒ OD = 2.5 m BD = 10 m ⇒ Coordinates of point B are (2.5, 10) Since the point B lies on the parabola $x^2$ = 4ay ∴ $(2.5)^2$ = 4a × 10 ⇒ a = $\frac{6.25}{40}$ = $\frac{625}{4000}$ = $\frac{5}{32}$ ∴ Equation of parabola is $x^2$ = 4 × $\frac{5}{32}$ y ⇒ $x^2$ = $\frac{5}{8}$ y Let PQ = d ⇒ NQ = $\frac{d}{2}$ ∴ Coordinates of point Q are $\left(\frac{d}{2},2\right)$ Since point Q lies on the parabola $x^2$ = $\frac{5}{8} y$ ∴ $\left(\frac{d}{2}\right)^2$ = $\frac{5}{8}$ × 2 ⇒ $\frac{d^2}{4}$ = $\frac{5}{4}$ ⇒ $d^2$ = 5 ⇒ d = $\sqrt{5}$ Thus width of arch = $\sqrt{5}$ m = 2.23 m approx.