A parabolic reflector with diameter PR = 20 cm and OQ = 5 cm. Vertex of the parabola is (0, 0). Let focus of the parabola be (a, 0) Now, PR = 20 cm ⇒ PQ = 10 cm ∴ Coordinate of point P are (5, 10) Since the point lies on the parabola $y^2$ = 4ax ∴ $(10)^2$ = 4 × 5 ⇒ a = $\frac{100}{20}$ ⇒ a = 5 Thus required focus of the parabola is (5, 0) which is the mid-point of the given diameter.