Diameter of the satellite dish

Question

A satellite dish is shaped like parabola with the receiver placed at the focus. It is to have a depth of 10. 25 m at the vertex, with the receiver placed 3.5 m away from the vertex. What should the diameter of the satellite dish be?

Answer 1

The approach is to first figure out the equation for a parabola satisfying

the constraint that its focus is 3.5 m from its vertex.

Once we have that, we can calculate anything about the parabola, such as

the diameter at 10.25 m from the vertex.

In general, a parabola has an equation of the form

f(x) = a(x - h)² + k

for some constants a, h, k. (Those constants determine its shape and position.)

The parabola's vertex is then at the point (h, k), and

the parabola's focus is at the point (h, k + 1/4a).

We have the freedom to place the parabola anywhere in the coordinate system.

The reason is that your question concerns only the parabola's shape, not its position.

The choice of position in a 2-dimentianal space allows us to fix 2 of the 3 parameters.

It is convenient to place its vertex in the origin.

Than means, we choose h = k = 0, so the parabola's equation becomes

f(x) = ax²

and its focus is then at the point (0, 1/4a).

There is just one parameter remaining -- a -- which controls the parabola's width.

We determine that from the requirement that the focus be 3.5 m from the vertex.

That means

1/4a = 3.5

Therefore

a = 1/(4×3.5) = 1/14

That gives us the parabola

f(x) = x²/14

Now we can determine the diameter at 10.25 m from the vertex.

Its radius, r, satisfies

f(r) = 10.25

That is because the radius is the distance form the y-axis of a point on the parabola

that is at height 10.25 from the vertex.

Solving the equation

r²/14 = 10.25

r = √(14×10.25) = 11.979

Therefore the diameter, being double the radius, is approximately 24 m.