1. A Parabolic Antenna.
A parabolic antenna has the shape of a paraboloid of revolution: it is created by rotating part of a parabola around its axis. By the reflection property, all signals arriving parallel to the axis are concentrated at the focus after reflecting off the dish, so that is where the receiver is mounted. The axial cross-section of the dish is described by two measurements: the diameter \(d\) of the dish and its depth \(h\text{.}\) Placing the vertex at the origin with the dish opening upward, the cross-section is the parabola \(x^2 = 4py\text{,}\) and the rim passes through the points \(\left(\pm\tfrac{d}{2},\, h\right)\text{.}\)
Consider a dish antenna with diameter \(d = 120\) cm and depth \(h = 20.3\) cm, suitable for the amateur radio band at \(5.76\) GHz.
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A parabola opening upward with vertex at the origin. The rim points at (minus d over 2, h) and (d over 2, h) are joined by a horizontal double arrow labeled d. A vertical double arrow labeled h shows the depth from the rim down to the axis level. The focus F(0,p) lies on the y axis above the rim, and two segments run from the focus to the rim points, enclosing the opening angle two phi.
(a)
Determine the optimal location for the receiver, that is, find the distance from the vertex of the dish to the focus.
Solution.
The receiver must sit at the focus \((0,p)\text{.}\) The rim point \(\left(\tfrac{d}{2}, h\right) = (60,\, 20.3)\) lies on the parabola \(x^2 = 4py\text{,}\) so
\begin{equation*}
60^2 = 4p(20.3)
\quad\Longrightarrow\quad
p = \frac{3600}{81.2} \approx 44.3.
\end{equation*}
The receiver should be mounted on the axis about \(44.3\) cm above the vertex of the dish. Note that \(p \gt h\) here, so the focus sits above the rim of this shallow dish.
(b)
Find the quadratic function \(y = f(x)\) (explicit form) that describes the curvature of the dish, and graph it (for example, in GeoGebra or Desmos).
Solution.
Solving \(x^2 = 4py\) for \(y\) with \(4p = \frac{3600}{20.3} \approx 177.3\) gives
\begin{equation*}
y = \frac{1}{4p}\,x^2 = \frac{20.3}{3600}\,x^2 \approx \frac{x^2}{177.3},
\qquad -60 \le x \le 60.
\end{equation*}
The graph captures the true curvature of the dish provided both axes use the same scale.
(c)
The opening angle \(2\varphi\) of the dish is the angle at which the two edges of the rim are seen from the focus. Compute it.
Solution.
The focus \(F(0,p)\text{,}\) the point \((0,h)\text{,}\) and the rim point \(\left(\tfrac{d}{2}, h\right)\) form a right triangle with legs \(p - h\) (vertical) and \(\tfrac{d}{2}\) (horizontal), and \(\varphi\) is the angle at the focus. Hence
\begin{equation*}
\tan\varphi = \frac{d/2}{p - h} = \frac{60}{44.3 - 20.3} = \frac{60}{24}
\quad\Longrightarrow\quad
\varphi \approx 68.2^\circ.
\end{equation*}
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The same parabola and focus as before, with a red right triangle drawn from the focus: a vertical leg of length p minus h down to the rim level, a horizontal leg of length d over 2 out to the right rim point, and the segment from the focus to the rim point as hypotenuse. The angle phi at the focus is marked.
The opening angle of the dish is \(2\varphi \approx 136.4^\circ\text{.}\)
