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Question 1168718: A flashlight is like a praboloid, so that if it's light bulb is placed at the focus, the light rays from the bulb will then bounce off the surface in a focused direction that is parallel to the axis. If the paraboloid has a depth 1.8 inches and the diameter on it's surface is 6 inches, how far should the light source be placed from the vertex
Answer by htmentor(1343)   (Show Source):
You can put this solution on YOUR website! Let's put the surface of the paraboloid along the x-axis, and center the paraboloid on the y-axis.
If the depth is 1.8 inches, then the vertex is at (0,-1.8)
The equation for a parabola which opens vertically can be written as:
(x-h)^2 = 4p(y-k), where (h,k) is the vertex.
We can use one of the two points [(-3,0) and (3,0)] at the edge of the surface to determine p.
x^2 = 4p(y+1.8) -> 3^2 = 4p(1.8) -> 9 = 7.2p -> p = 1.25
The focus of the parabola is located at:
F = (h,k+p), thus the light source should be placed at (0,-1.8+1.25) -> (0,-0.55)
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