a cross-section with the vertex at (0,0) shows that if y = ax^2, y(15) = 10.
So, a = 10/225 = 2/45
y = 2/45 x^2
Now recall that the parabola
x^2 = 4py has its focus at (0,p)
So, with x^2 = 45/2 y, p = 45/8
The receiver should be 45/8 inches from the vertex.
a satellite dish is the shape of a paraboloid. The dish is 30 inches wide, and 10 inches deep. How many inches should the receiver be located from the vertex for optimal reception (round to the nearest thousandth)
1 answer
1 answer