Chapter 9.1, Problem 96E

To determine

To find the equation of the path taken by water

Answer to Problem 96E

The equation that describes the parabola is

  $-4x^2=-25\left(y-48\right)$

Explanation of Solution

Given information :

Water flows from horizontal pipe 48 feet above the ocean surface.

Water follows parabolic path having vertex $\left(0,48\right)$ at the end of the pipe.

The point at which water strikes the ocean is $\left(10\sqrt{3},0\right)$ .

  

Concept used:

The standard equation of a parabola that has vertical axis of symmetry is:

  ${\left(x-h\right)}^2=4p\left(y-k\right)\left(where\left(h,k\right)\text{ is vertex of parabola}\right)$

From the given information, vertex is located at the coordinates $\left(0,48\right)$ and the endpoint is at coordinates $\left(10\sqrt{3},0\right)$

From above clearly seen

  $\left(h,k\right)\to (0,48)\text{ and }\left(x,y\right)\to \left(10\sqrt{3},0\right)$

Substituting values of vertex and endpoints in standard equation of parabola to find value of p

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

Getting equation of parabola by putting values of $\left(h,k\right)$ and $p$ in standard equation of parabola

  $\begin{array}{l}{\left(x-0\right)}^2=4\left(\frac{-25}{16}\right)\left(y-48\right)\\ x^2=\frac{-25}{4}\left(y-48\right)\\ 4x^2=-25\left(y-48\right)\end{array}$

Therefore,

The equation that describes the parabola is

  $-4x^2=-25\left(y-48\right)$