Chapter 11.1, Problem 54E

Satellite Dish A reflector for a satellite dish is parabolic in cross section, with the receiver at the focus F. The reflector is 1 ft deep and 20 ft wide from rim to rim (see the figure). How far is the receiver from the vertex of the parabolic reflector?

To determine

To find: The distance of the receiver from the vertex, located at the focus of the parabolic reflector with $1$ ft deep and $20$ ft wide.

Answer to Problem 54E

The distance of the receiver at the focus from the vertex of the parabola is $25\text{ft}$.

Explanation of Solution

Definition used:

The equation of the parabola with vertex $V(0,0)$, focus $F\left(0,p\right)$ and directrix $y=-p$ is $x^2=4py$.

Calculation:

From the given diagram observe that the parabola is open upward.

Thus, the equation of the parabola is of the form $x^2=4py$.

Given that the reflector is 1ft deep, 20m wide and the axis divides the 20ft wide in to two halves.

Let $V(0,0)$ be the vertex of the reflector of 1ft deep.

Therefore, the point $\left(0,1\right)$ is on the positive y –axis.

When y =1 there are two points on the parabola.

Substitute y =1 in $x^2=4py$,

$\begin{array}{l}{x}^2=4py\\ x^2=4p\left(1\right)\\ x=\pm 2\sqrt{p}\end{array}$

The focal diameter is 4p.

Given that the width of the dish is $20m$.

Substitute $20m$ in $x=\pm 2\sqrt{p}$,

$\begin{array}{l}x=\pm 2\sqrt{p}\\ 2x=4\sqrt{p}\\ 20=4\sqrt{p}\\ \sqrt{p}=5\end{array}$

$p=25$

Thus, focus of the parabola reflector is $F\left(0,25\right)$

Therefore, the receiver is at a distance of $25\text{ft}$ from the reflector.