Chapter 9, Problem 25RE

To determine

Find the width of the archway at ground level.

Answer to Problem 25RE

The width of the archway is $8\sqrt{6}meters$ .

Explanation of Solution

Given:

It is given in the question that a parabolic archway is $12$ meters high at the vertex. At a height of $10$ meters, the width of the archway is $8$ meters.

Concept Used:

In this, use the concept of the parabola and use the formula is ${(x-h)}^2=4p(y-k)$ .

Calculation: Here, the parabola is a vertical axis, so ${(x-h)}^2=4p(y-k)$ will use.

Now, use the vertex and one of the given points to solve for p.

Let, the vertex be $(0,12)$ and the given point be $(4,10)$

So,

  $\begin{array}{l}$ {(x-h)}^2=4p(y-k)$$ {(4-0)}^2=4p(10-12)$16=4p\times -2\\ p=\frac{16}{4\times -2}=-2\end{array}$

Now, use p and the vertex to write the equation for the parabola. Let $y=0$ , Solve for the positive value of x. The width at ground level is $2x$ .

  $\begin{array}{l}$ {(x-h)}^2=4p(y-k)$$ {(x-0)}^2=4(-2)(0-12)$x^2=-8\times -12\\ x^2=96\\ x=\sqrt{96}=4\sqrt{6}\\ Here,2x=2\times 4\sqrt{6}=8\sqrt{6}\end{array}$

Conclusion:

The width is $8\sqrt{6}meters$ .