Question

suppose that a satellite receiver with a parabolic cross - section is 42 inches across and 12 inches deep. how far from the vertex must the receptor unit be located to ensure that it is at the focus of the parabola? inches

Explanation:

Step1: Set up the equation of the parabola

Assume the parabola is of the form \(y = ax^{2}\), and it passes through the point \((21,12)\) (since the parabola is 42 inches across, \(x = 21\) when \(y = 12\)). Substitute \(x = 21\) and \(y = 12\) into \(y=ax^{2}\), we get \(12=a\times(21)^{2}\).

Step2: Solve for \(a\)

From \(12 = 441a\), we can find \(a=\frac{12}{441}=\frac{4}{147}\).

Step3: Use the formula for the focus of a parabola \(y = ax^{2}\)

The formula for the focus of the parabola \(y = ax^{2}\) is \((0,\frac{1}{4a})\). Substitute \(a=\frac{4}{147}\) into \(\frac{1}{4a}\), we have \(\frac{1}{4\times\frac{4}{147}}=\frac{147}{16}=9.1875\).

Answer:

\(9.1875\)