|  | 
| 
 
 
| Question 1198216:  A satellite dish is shaped like a paraboloid of revolution. This means that it can be formed by rotating a parabola around its axis of symmetry. The receiver is to be located at the focus. If the dish is 72 feet across at its opening and 6 feet deep at its center, where should the receiver be placed? (Hint: Draw a cross section of the dish on a graph and place the vertex at (0, -6) so that the opening of the dish lies on the x-axis).
 Find the equation of the parabola.
 How far above the vertex should the receiver be placed?
 Answer by onyulee(41)
      (Show Source): 
You can put this solution on YOUR website! **1. Find the Equation of the Parabola** * **Vertex:** (0, -6)
 * **Point on the Parabola:** (36, 0) (Since the dish is 72 feet across, the point on the parabola is half of that, or 36 feet, from the vertex)
 * **Standard Form of a Parabola:**
 (x - h)² = 4p(y - k)
 where (h, k) is the vertex
 * **Substitute values:**
 (x - 0)² = 4p(y - (-6))
 x² = 4p(y + 6)
 * **Find the value of 'p':**
 Substitute the point (36, 0) into the equation:
 36² = 4p(0 + 6)
 1296 = 24p
 p = 1296 / 24
 p = 54
 * **Equation of the Parabola:**
 x² = 4 * 54 * (y + 6)
 x² = 216(y + 6)
 **2. Find the Distance of the Receiver from the Vertex**
 * The receiver should be placed at the focus of the parabola.
 * The distance from the vertex to the focus is 'p'.
 * **Receiver Distance:** 54 feet
 **Therefore:**
 * The equation of the parabola is: x² = 216(y + 6)
 * The receiver should be placed 54 feet above the vertex of the dish.
 
 | 
  
 | 
 |  |  |