Here, width of elliptical arch = 8 m ∴ AB = 8 m ⇒ 2a = 8 ⇒ a = 4 Height at the centre = 2 m ∴ OC = 2 ⇒ b = 2 The axis of the ellipse is x-axis. So the equation of ellipse in standard form is $\frac{x^2}{a^2} + \frac{y^2}{b^2}$ = 1. ∴ $\frac{x^2}{(4)^2} + \frac{y^2}{(2)^2}$ = 1 ⇒ $\frac{x^2}{16} + \frac{y^2}{4}$ = 1 Now, AP = 1.5 m OP = OA – AP = 4 – 1.5 = 2.5 m Let PQ = h m ∴ Coordinates of Q are (2.5, h) Since the point Q lies on the ellipse $\frac{x^2}{16} + \frac{y^2}{4}$ = 1 ∴ $\frac{(2.5)^2}{16} + \frac{h^2}{4}$ = 1 ⇒ $\frac{h^2}{4}$ = $\frac{16-6.25}{16}$ ⇒ $h^2$ = $\frac{9.75 \times 4}{16}$ = $\frac{9.75}{4}$ ⇒ $h^2$ = 2.44 ⇒ h = $\sqrt{2.44}$ = 1.56 m approx.