Chapter 9, Problem 26STP

To determine

To find: that how wide is archway at the ground level.

Answer to Problem 26STP

The archway is $\text{33}\text{.94m}$ wide at the ground level.

Explanation of Solution

Given information:

From the figure it can be observed that parabola vertex is $\left(0,16\right)$ and points on parabola is $\left(6,14\right)$ .

Calculation: to find the how wide is archway at the ground level, first find the standard form of parabola,

Since, the axis of parabola is vertical then the standard form of parabola is ${\left(x-h\right)}^2=4p\left(y-k\right)$ .

Since, vertex is $\left(0,16\right)$ so $h=0,k=16$ .

Thus, the equation is $x^2=4p\left(y-16\right)$ .

Now, the point at parabola is $\left(6,14\right)$ , so substitute in the equation,

  $\begin{array}{l}{x}^2=4p\left(y-16\right)\\ {\left(6\right)}^2=4p\left(14-16\right)\text{ [substitute]}\\ 36=4p\left(-2\right)\text{ [subtract and multiply]}\\ p=-4.5\end{array}$

Now the equation of parabola is $x^2=-18\left(y-16\right)$ .

Now, to find the wideness of archway at the ground level, substitute $y=0$ in the above equation and simplify,

  $\begin{array}{l}{x}^2=-18\left(y-16\right)\\ x^2=-18\left(0-16\right)\text{ [substitute]}\\ x^2=288\text{ [multiply]}\\ x=16.97\text{ [simplify]}\end{array}$

Thus, The archway is $\text{33}\text{.94m}$ wide at the ground level.