Chapter 10.2, Problem 58PPSE

Solution

To Show: The radius is four times the focal length.

It has been shown that the radius is four times the focal length.

Given:

The radius and the depth of a satellite dish are equal.

Concept used:

The equation of the vertical parabola with vertex at origin is $y=\frac{1}{4c}{x}^2$ .

The focus of this parabola is at $\left(0,c\right)$ .

Calculation:

Let the satellite dish be oriented such that it forms a vertical parabola with vertex at origin.

It is given that the radius and the depth of the satellite dish are equal.

Then, the end points of the diameter are of the form $\left(a,b\right)$ and $\left(-a,b\right)$ .

The radius is then $a$ and the depth is then $b$ .

According to the problem,

  $b=a$

Put $b=a$ in $\left(a,b\right)$ to get $\left(a,a\right)$ .

Now, the point $\left(a,a\right)$ lies on the parabola.

Put $\left(x,y\right)=\left(a,a\right)$ in $y=\frac{1}{4c}{x}^2$ to get,

  $\begin{array}{c}a=\frac{1}{4c}{a}^2\\ 1=\frac{1}{4c}a\\ c=\frac{1}{4}a\end{array}$

So, $c=\frac{1}{4}a$ .

Now, the focus is at $\left(0,c\right)=\left(0,\frac{1}{4}a\right)$ .

Now, the focal length is the distance between the vertex, $\left(0,0\right)$ and the focus, $\left(0,\frac{1}{4}a\right)$ , which is $\frac{1}{4}a$ .

It has been assumed that the radius is $a$ .

Comparing, it follows that the radius is four times the focal length.

Conclusion:

It has been shown that the radius is four times the focal length.