|
Question 1170538: A satellite dish is shaped like a paraboloid, with the receiver placed at the focus. It is to have a depth of 0.44 m at the vertex, with the receiver placed 0.11 m away from the vertex. What should the diameter of
the satellite dish be?
Answer by CPhill(1959)   (Show Source):
You can put this solution on YOUR website! Let the paraboloid be represented by the equation $x^2 = 4py$, where the vertex is at the origin (0, 0) and the focus is at (0, p).
We are given that the receiver is placed 0.11 m away from the vertex, which means the focus is at (0, 0.11). Therefore, p = 0.11 m.
The equation of the paraboloid is $x^2 = 4(0.11)y$, or $x^2 = 0.44y$.
The depth of the dish is 0.44 m, which means when y = 0.44 m, we can find the x-coordinate of the edge of the dish.
Substitute y = 0.44 into the equation:
$x^2 = 0.44(0.44) = 0.1936$
$x = \pm \sqrt{0.1936} = \pm 0.44$
The diameter of the dish is the distance between the two x-coordinates, which is:
Diameter = 0.44 - (-0.44) = 2(0.44) = 0.88 m.
Therefore, the diameter of the satellite dish should be 0.88 m.
Final Answer: The final answer is $\boxed{0.88}$
|
|
|
| |
