Using your dimensions of the parabola, let's say that it is of the form
y = ax^2
Since y(30)=30, a = 1/30
y = 1/30 x^2
Now recall that the parabola
x^2 = 4py
has focus at (0,p). So, 4p=30 and p = 7/2. So the focus should be 3.5 inches from the vertex.
The reflecting dish of a parabolic microphone has a cross-section in the shape of a parabola. The microphone itself is placed on the focus of the parabola. If the parabola is 60 inches wide and 30 inches deep, how far from the vertex should the microphone be placed?
1 answer