Question

a satellite dish is shaped like a paraboloid of revolution. this means that it can be formed by rotating a parabola around its axis of symmetry. the receiver is to be located at the focus. if the dish is 48 feet across at its opening and 6 feet deep at its center, where should the receiver be placed?
find the equation of the parabola.
how far above the vertex should the receiver be placed?

Explanation:

Step1: Define the parabola's vertex and point

We place the vertex of the parabola at the origin \((0,0)\) and open it upwards, so the standard form of the parabola is \(x^{2}=4py\). The dish is 48 feet across at the opening and 6 feet deep, so when \(y = 6\), \(x=\pm24\) (since half of 48 is 24).

Step2: Substitute the point into the equation

Substitute \(x = 24\) and \(y=6\) into \(x^{2}=4py\):

$$
LATEXBLOCK0
$$

Step3: Solve for \(p\)

Divide both sides by 24: \(p=\frac{576}{24} = 24\)? Wait, no, wait, let's recalculate. Wait, \(24^{2}=576\), \(4p\times6 = 24p\), so \(24p=576\), then \(p=\frac{576}{24}=24\)? Wait, no, that can't be. Wait, no, the standard form for a parabola opening upwards with vertex at \((0,0)\) is \(x^{2}=4py\), where \(p\) is the distance from the vertex to the focus. Wait, let's check again. When \(y = 6\), \(x = 24\) (since the diameter is 48, radius 24). So plug into \(x^{2}=4py\): \(24^{2}=4p\times6\), so \(576 = 24p\), so \(p=\frac{576}{24}=24\)? Wait, that seems large, but let's check the equation of the parabola first. Wait, no, maybe I made a mistake. Wait, the depth is 6 feet, so at \(y = 6\), \(x = 24\). So the equation of the parabola: from \(x^{2}=4py\), substituting \(x = 24\), \(y = 6\), we get \(24^{2}=4p\times6\), so \(576=24p\), so \(p = 24\)? Wait, no, 24 times 24 is 576, 4p times 6 is 24p, so 24p=576, so p = 24/1? Wait, no, 576 divided by 24 is 24? Wait, 24 times 24 is 576, yes. Wait, but then the focus is at \((0,p)=(0,24)\)? But that seems too deep. Wait, no, maybe the vertex is at the bottom, so the parabola opens upwards, with vertex at \((0,0)\), and when \(y = 6\), \(x=\pm24\). So the equation is \(x^{2}=4py\). Then solving for \(p\): \(p=\frac{x^{2}}{4y}=\frac{24^{2}}{4\times6}=\frac{576}{24}=24\). Wait, but that would mean the focus is 24 feet above the vertex. But let's check the equation of the parabola. So the equation of the parabola is \(x^{2}=4\times24y=96y\), so \(x^{2}=96y\). Wait, but let's verify. If \(y = 6\), then \(x^{2}=96\times6 = 576\), so \(x=\pm24\), which matches. Then the focus is at \((0,p)=(0,24)\)? Wait, no, wait, the standard form is \(x^{2}=4py\), so the focus is at \((0,p)\). So if \(p = 24\), then the focus is 24 feet above the vertex. But that seems very deep, but maybe it's correct. Wait, maybe I messed up the standard form. Wait, another way: the standard form for a parabola with vertex at \((h,k)\) is \((x - h)^{2}=4p(y - k)\). Here, \(h = 0\), \(k = 0\), so \((x - 0)^{2}=4p(y - 0)\), so \(x^{2}=4py\). So when \(y = 6\), \(x = 24\), so \(24^{2}=4p\times6\), so \(576 = 24p\), so \(p = 24\). So the equation of the parabola is \(x^{2}=96y\), and the receiver (focus) is 24 feet above the vertex. Wait, but that seems like a lot, but let's check with a smaller example. Suppose a parabola 4 feet across (x=2) and 1 foot deep (y=1). Then \(x^{2}=4py\), so \(4 = 4p\times1\), so \(p = 1\), which makes sense, focus at (0,1), 1 foot above vertex. So in our case, 48 feet across (x=24) and 6 feet deep (y=6), so \(24^{2}=4p\times6\), so \(576 = 24p\), so \(p = 24\). So that seems correct.

Answer:

The equation of the parabola is \(x^{2}=96y\), and the receiver should be placed 24 feet above the vertex. Wait, no, wait, wait, I think I made a mistake. Wait, 24 squared is 576, 4p times 6 is 24p, so 24p=576, so p=24? Wait, 576 divided by 24 is 24? Yes, 24 times 24 is 576. So the focus is at (0,24), so 24 feet above the vertex. And the equation of the parabola is \(x^{2}=96y\) (since 4p=96 when p=24).