A satellite dish in the shape of a paraboloid opening upwards is 10 feet across, and 4 feet deep at its
vertex. How far is the receiver from the vertex, if it is placed at the focus? Round off your answer to the
nearest hundredths. Support your answer with a corresponding illustration

To solve this problem, we need to determine the location of the focus of the paraboloid, as this is where the receiver will be placed. The focus of a paraboloid is determined by the shape of its cross-section, which is a parabola.

### Step-by-step Solution:

1. **Understanding the Geometry of the Paraboloid:**

   - The paraboloid opens upwards, indicating its cross-section is a parabola that opens upwards.
   - The paraboloid is 10 feet across at the opening, which means the diameter of the parabola's top is 10 feet. Thus, the radius of the parabola is \(5\) feet.
   - The depth of the paraboloid at the vertex is \(4\) feet.

2. **Equation of the Parabola:**

   Since the paraboloid is symmetric and opens upward, the parabola's vertex can be placed at the origin of a coordinate system. The standard form of a parabola that opens upwards with its vertex at \((0, 0)\) is given by:

   \[
   y = ax^2
   \]

   where \( a \) is a constant that determines the parabola's shape.

3. **Using Given Points to Determine \(a\):**

   We know that the parabola passes through the point \((5, 4)\) since it is 5 feet from the center (half the diameter) and 4 feet deep at the vertex. Plugging this point into the parabola equation:

   \[
   4 = a \cdot 5^2
   \]

   \[
   4 = 25a
   \]

   Solving for \( a \):

   \[
   a = \frac{4}{25}
   \]

4. **Finding the Focus of the Parabola:**

   The distance from the vertex to the focus \( f \) of a parabola \( y = ax^2 \) is given by:

   \[
   f = \frac{1}{4a}
   \]

   Substituting the value of \( a \):

   \[
   f = \frac{1}{4 \cdot \frac{4}{25}}
   \]

   \[
   f = \frac{1}{\frac{16}{25}}
   \]

   \[
   f = \frac{25}{16}
   \]

   \[
   f = 1.5625 \text{ feet}
   \]

   So, the receiver should be placed approximately **1.56 feet** from the vertex of the paraboloid.

### Illustration:

To help visualize, imagine the parabola with its vertex at the origin (0, 0) in a 2D coordinate system:

1. Draw a parabola opening upwards.
2. The vertex of the parabola is at the origin (0,0).
3. The point (5, 4) lies on the parabola, 5 units horizontally from the vertex and 4 units vertically up.
4. The focus is positioned 1.56 feet above the vertex, along the axis of symmetry.

### Summary

- **Distance of the receiver from the vertex (focus):** Approximately **1.56 feet**.
  
Would you like further details on this problem, or do you have any other questions?

### Additional Questions for Exploration:

1. How would the location of the focus change if the depth of the paraboloid were different?
2. What is the significance of the focus point in the functioning of a satellite dish?
3. If the satellite dish were twice as wide, how would the focus's position change?
4. How would you derive the focus for a parabolic dish with a different cross-sectional shape, like an ellipse?
5. What is the relationship between the depth and the width of a parabolic dish in determining its focus?

### Tip:

Understanding how to derive the focus of a parabola from its geometric properties can be extremely helpful in applications like satellite dishes, telescopes, and even headlight reflectors, as these all rely on parabolic shapes to direct signals and light effectively.

Learn how to calculate the distance of a receiver from the vertex of a paraboloid-shaped satellite dish. This problem involves understanding parabolas, focusing on their geometric properties, and applying the formula for determining the focus distance. Suitable for high school students studying geometry.

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