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cmboutin27 · Apr 28, 2010, 10:28 AM

If a parabolic satellite has a diameter of 6 feet and depth of 2 feet, how far is the focus from the vertex?

I have absolutely no idea where to begin!

ebaines · Apr 28, 2010, 01:37 PM

Consider what happens if you slice the parabollic antenna across the middle, giving us a 2-dimensional edge in the shape of a parabola. You can define any parabola if you know its vertex and any other point on the curve. Here you know that the center of the parabola (its vertex) is 2 feet below the outside edge, and the outside edge is 3 feet to the left and right of the center. So you can say that the center of the parabola is at (0,0), the right edge is at (3,2), and the left edge is at (-3,2)

The standard form of the parabolic equation that you want to use here is:

$ (x-x_1)^2 = 4p(y-y_1) $

where $(x_1, y_1)$ is the vertex and $p$ is the distance from vertex to focus. The value of $p$ is what you need to solve for.

You know that $(x_1, y_1) = (0,0)$ , so this formula becomes:

$ (x-0)^2 = 4p(y-0), \\ y = \frac {x^2} {4p} $

So now you need to find the value of $p$ . Given that the point (3,2) is on this parabola, simply use that point to determine p:

$ 2 = \frac {3^2} {4p} $

I'm sure you can take it from here!