Chapter 9.1, Problem 99E

To determine

Tofind:The equation of the path taken by the water.

Answer to Problem 99E

The equation of the path taken by the water is $4x^2=25\left(48-y\right)$ .

Explanation of Solution

Given information:

Water is flowing from a horizontal pipe $48$ feet above the ground. The falling stream of water has the shape of a parabola whose vertex $\left(0,48\right)$ is at the end of the pipe. The stream of water strikes the ocean at the point $\left(10\sqrt{3},0\right)$ .

  

Calculation:

The standard equation for the parabola is ${\left(x-h\right)}^2=4p\left(y-k\right)$ .

As clear from the figure that the parabola is vertically downwards with vertex at $\left(0,48\right)$ .

Now substitute $0$ for $h$ and $8$ for $k$ in the general equation.

  $\begin{array}{c}{\left(x-0\right)}^2=4p\left(y-48\right)\\ x^2=4p\left(y-48\right)\end{array}$

Now as it is clear from the figure that the parabola passes through the point $\left(10\sqrt{3},0\right)$ .

Now substitute $10\sqrt{3}$ for $x$ and $0$ for $y$ in the general equation.

  $\begin{array}{c}{x}^2=4p\left(y-48\right)\\ {\left(10\sqrt{3}\right)}^2=4p\left(0-48\right)\\ 300=-192p\\ p=-\frac{25}{16}\end{array}$

Now substitute $-\frac{25}{16}$ for $p$ in the equation of parabola.

  $\begin{array}{l}{x}^2=4p\left(y-48\right)\\ x^2=4\left(-\frac{25}{16}\right)\left(y-48\right)\\ x^2=-\frac{25}{4}\left(y-48\right)\\ 4x^2=25\left(48-y\right)\end{array}$

Therefore, the equation of the path taken by the water is $4x^2=25\left(48-y\right)$ .