SOLUTION: Radiation is focused to an unhealthy area in a patient's body using a parabolic reflector, positioned in such a way that the target area is at the focus. If the reflector is 30cm w

Algebra ->  Quadratic-relations-and-conic-sections -> SOLUTION: Radiation is focused to an unhealthy area in a patient's body using a parabolic reflector, positioned in such a way that the target area is at the focus. If the reflector is 30cm w      Log On


   



Question 1196496: Radiation is focused to an unhealthy area in a patient's body using a parabolic reflector, positioned in such a way that the target area is at the focus. If the reflector is 30cm wide and 15 cm deep at the center, how far should the base of the reflector be from the target area?
Answer by MathLover1(20850) About Me  (Show Source):
You can put this solution on YOUR website!

let vertex be at origin, V(0,0)
if the reflector is 30cm wide and 15cm deep, we have two points that lie on parabola and these are:
(15,15) and (15,-15) (means parabola opens to the right)

general formula for this parabola is:
y%5E2=4px
The value of p is the distance from the vertex to the focus of the parabola. A focus always lies on the axis of its parabola.

use one point to calculate p
15%5E2=4p%2A15........simplify
15=4p
p=15%2F4
p=3.75
your parabola is:
y%5E2=4%2A3.75x
y%5E2=15x
since the p+=+3.75 means the focus is 3.75+unit to the right of the vertex at coordinate (3.75, 0)
so, the base should be 3.75cm from the target