Chapter 9.1, Problem 98E

To determine

Tofind:The width of the lattice arch at the ground level.

Answer to Problem 98E

The width of lattice arch at ground level is $4\sqrt{2}$ .

Explanation of Solution

Given information:

A parabolic arch is $8$ feet high at the vertex. At a height of $4$ feet, the width of the lattice arch is $4$ feet.

  

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 $\left(0,8\right)$ .

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

  $\begin{array}{c}{\left(x-0\right)}^2=4p\left(y-8\right)\\ x^2=4p\left(y-8\right)\end{array}$

Now as it is clear from the figure that the parabola passes through the point $\left(2,4\right)$ .

Now substitute $2$ for $x$ and $4$ for $y$ in the general equation.

  $\begin{array}{c}{x}^2=4p\left(y-8\right)\\ 2^2=4p\left(4-8\right)\\ 4=-16p\\ p=-\frac{1}{4}\end{array}$

Now substitute $-\frac{1}{4}$ for $p$ in the equation of parabola.

  $\begin{array}{l}{x}^2=4p\left(y-8\right)\\ x^2=4\left(-\frac{1}{4}\right)\left(y-8\right)\\ x^2=-\left(y-8\right)\\ x^2=8-y\end{array}$

Now to find the x-intercepts substitute $0$ for $y$ in the above equation.

  $\begin{array}{c}{x}^2=8-y\\ x^2=8\\ x=\pm 2\sqrt{2}\end{array}$

So, the distance between both the x-intercepts is $4\sqrt{2}$ .

Therefore, the width of lattice arch at ground levelis $4\sqrt{2}$ .