A searchlight has a parabolic reflector that forms a "bowl," which is 18 inches wide from rim to rim and 10 inches deep. If the filament of the lightbulb is located at the focus, how far from the vertex of the reflector is it? Round your answer to the nearest hundredth. You may want to draw a graph to help you with this problem. Include it as part of your answer.

Answer 1

We can start by using the standard equation for a parabola with vertex at the origin:

y = 1/(4p) x^2

where p is the distance from the vertex to the focus. We want to find p, so we need to first find the equation of the parabola in terms of x and y.

Since the bowl is 18 inches wide, we know that the distance between the two points on the parabola that are 9 inches from the vertex in the x-direction is also 18 inches. This means that the x-coordinate of each of those points is ±9. Plugging this into the equation of the parabola gives us:

y = 1/(4p) (±9)^2 = 81/(4p)

Simplifying this gives us:

4py = 81

Now we need to use the fact that the bowl is 10 inches deep. This means that the y-coordinate of the point on the parabola that is 5 inches above the vertex is 10. Plugging this into the equation of the parabola gives us:

10 = 81/(4p) + c

where c is the y-coordinate of the vertex. But we know that the vertex is at (0, -10), so c = -10. Plugging this in and simplifying gives us:

40p = 71

So p = 71/40, which is approximately 1.78 inches. Therefore, the filament of the lightbulb is located 1.78 inches from the vertex of the reflector.

Here is a graph to help visualize the situation: