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# A satellite dish has a diameter of 8 feet. The depth of the dish is 1 foot at the center of the dish. Where should the receiver be placed?

$\begin{array}{1 1} 4 \; ft\; above\; the\; vertex \\ 4 \; ft \;below \;the\; vertex \\ 2 \; ft\; above\; the\; vertex \\ At\; at\; 60^{\circ} angle\;to\; the\; satellite \end{array}$

The problem can be represented as a parabola as follows, by placing the dish at Vertex $(0,0)$
We can observe that point $A \;=\; (4,1)$.
Since the equation of the parabola is $y = ax^2 \rightarrow$ at point $A, y = 1 = a \ times (4)^2 \rightarrow a = \large\frac{1}{16}$
The received should be placed at the focus of the parabola.
Given the equation of the parabola, the distance to the focus = $\large\frac{1}{4a}$ $= \large\frac{1 \times 16}{4 \times 1}$$= 4 \;ft$
The receiver should be placed $4\;ft$ above the vertex.