A cross section of a parabolic reflector has a vertical axis of symmetry with its vertex at (0,0).

Question

A cross section of a parabolic reflector has a vertical axis of symmetry with its vertex at (0,0). The focus of the reflector is 5 meters above the vertex. The reflector extends 3 feet to either side of the vertex. What is the depth of the reflector? Round your answer to the nearest hundredth.

Answer 1

It sounds like the depth is 6'. Ineed to see a graph.

Answer 2

With a vertical axis of symmetry, we have the equation of the parabola as

y = a(x-h)² + k

where (h,k) is the vertex. The vertex is given as (0,0), so h = 0 and k = 0. Therefore, our equation looks like this

y = a(x-0)² + 0

y = ax²

The focus is 5 meters above the vertex, so the focus is at (0,5). The focus is given by (h,k + 1/4a). With this we can solve for a as follows:

k + 1/4a = 5

0 + 1/4a = 5

1/4a = 5

1/(4)(5) = a

a = 1/20

Putting this back into our equation we get this

y = a(x-h)² + k

y = (1/20)(x-0)² + 0

y = (1/20)x²

The depth, y, of the reflector can be solved by putting in the x value from the vertex. Here, x is either -3 or 3:

y = (1/20)(3)²

y = 9/20

Therefore, the depth is 9/20 meters. *NOTE: I am assuming that the distance 3 is suppose to be in meters and not feet. Remember that both the focus and the distance measured from the vertex must be in the same units!

Hope this helps!