SOLUTION: A domed ceiling is a parabolic surface. For the best lighting on the floor, a light source is to be placed at the focus of the surface. If 1m down from the top of the dome the ceil

Algebra ->  Quadratic-relations-and-conic-sections -> SOLUTION: A domed ceiling is a parabolic surface. For the best lighting on the floor, a light source is to be placed at the focus of the surface. If 1m down from the top of the dome the ceil      Log On


   



Question 458923: A domed ceiling is a parabolic surface. For the best lighting on the floor, a light source is to be placed at the focus of the surface. If 1m down from the top of the dome the ceiling is 17m wide, find the best location for the light source.
Thank you for the help.

Answer by lwsshak3(11628) About Me  (Show Source):
You can put this solution on YOUR website!
A domed ceiling is a parabolic surface. For the best lighting on the floor, a light source is to be placed at the focus of the surface. If 1m down from the top of the dome the ceiling is 17m wide, find the best location for the light source.
..
Standard form of parabola which applies to this problem: (x-h)^2=4p(y-k), with (h,k) being the (x,y) coordinates of the vertex and p=distance from vertex to focal point. To solve given problem, we must find p.
..
Place the left side of the parabola at the origin (0,0), and the right side at(17,0).
This places the vertex at (8.5,1)=(h,k)
(x-8.5)^2=4p(y-1)
plug in (x,y) coordinates from one of given points on parabola (try (17,0))to solve for p
(17-8.5)^2=4p(0-1)
(8.5)^2=-4p
72.25=-4p
p=72.25/-4=|-18|=18 m (rounded)
ans:
The best location for the light source is 18 m below the top of the dome or at (8.5,-17)