Question
A cross-section of a parabolic reflector and bulb is located at the focus is 16cm.find the equation of the parabola and the diameter of the opening (R,S),11cm from the vertex
Answers
Steve
Your reflector is just a parabola revolved about its axis.
Recall that the parabola
x^2 = 4py has
latus rectum = 2p
focus at (0,p)
vertex at (0,0)
So, your parabola has p=8, making its equation
x^2 = 32y
Now you want 2x where y=11
That will be 2√(32*11) = 8√22
Recall that the parabola
x^2 = 4py has
latus rectum = 2p
focus at (0,p)
vertex at (0,0)
So, your parabola has p=8, making its equation
x^2 = 32y
Now you want 2x where y=11
That will be 2√(32*11) = 8√22