At a water fountain, water attains a maximum height of 4 m at horizontal distance of 0.5 m from its origin. If the path of water is a parabola, find the height of water at a horizontal distance of 0.75 m from the point of origin.

#### Solution

Let the equation of parabola be

(x – h)^{2} = – 4a(y – k)

Here vertex B(0.5, 4)

∴ Equation of parabola

(x – 0.5)^{2} = – 4a(y – 4)

Parabola passes through origin (0, 0)

(0 – 0.5)^{2} = – 4a(0 – 4)

`(- 1/2)^2` = 16a

∴ `1/4` = 16a

⇒ a =`1/64`

∴ Equation of parabola

(x – 0.5)^{2} = `- 4 (1/64)(y - 4)`

This Parabola passes again through D(0.75, y_{1})

∴ (0.75 – 0.5)^{2} = `- 1/16 (y_1 - 4)`

(0.25)^{2 }= `- 1/16 (y_1 - 4)`

`(1/4)^2 = - 1/16 (y_1 - 4)`

`1/16 = - 1/16 (y_1 - 4)`

1 = – y_{1} + 4

y_{1} = 3

Height of water at a horizontal distance of 0.75 m is 3 m.