find an equation of the surface consisting of all points P(x,y,z) that are twice as far from the plane z=-4 as from the point (0,0,4). identify the surface

We are tasked with finding an equation of a surface that consists of all points \( P(x, y, z) \) such that the distance from \( P \) to the plane \( z = -4 \) is twice the distance from \( P \) to the point \( (0, 0, 4) \).

### Step 1: Distance from \( P(x, y, z) \) to the plane \( z = -4 \)
The formula for the distance from a point \( P(x, y, z) \) to the plane \( z = c \) is given by the absolute difference between \( z \) and \( c \). Therefore, the distance from \( P(x, y, z) \) to the plane \( z = -4 \) is:

\[
\text{Distance to the plane} = |z - (-4)| = |z + 4|
\]

### Step 2: Distance from \( P(x, y, z) \) to the point \( (0, 0, 4) \)
The distance from a point \( P(x, y, z) \) to another point \( (0, 0, 4) \) is given by the Euclidean distance formula:

\[
\text{Distance to the point} = \sqrt{x^2 + y^2 + (z - 4)^2}
\]

### Step 3: Set up the relationship
We are told that the distance from \( P \) to the plane \( z = -4 \) is twice the distance from \( P \) to the point \( (0, 0, 4) \). This gives the equation:

\[
|z + 4| = 2 \cdot \sqrt{x^2 + y^2 + (z - 4)^2}
\]

### Step 4: Square both sides to eliminate the square root
Squaring both sides of the equation:

\[
(z + 4)^2 = 4 \cdot (x^2 + y^2 + (z - 4)^2)
\]

### Step 5: Expand both sides
Expanding the left-hand side:

\[
(z + 4)^2 = z^2 + 8z + 16
\]

Expanding the right-hand side:

\[
4 \cdot (x^2 + y^2 + (z - 4)^2) = 4(x^2 + y^2 + z^2 - 8z + 16)
\]
\[
= 4x^2 + 4y^2 + 4z^2 - 32z + 64
\]

### Step 6: Simplify the equation
Now, equate the two expanded sides:

\[
z^2 + 8z + 16 = 4x^2 + 4y^2 + 4z^2 - 32z + 64
\]

Move all terms to one side of the equation:

\[
z^2 + 8z + 16 - 4x^2 - 4y^2 - 4z^2 + 32z - 64 = 0
\]

Simplify:

\[
-3z^2 + 40z - 48 - 4x^2 - 4y^2 = 0
\]

Divide by -3 to simplify:

\[
z^2 - \frac{40}{3}z + 16 + \frac{4}{3}(x^2 + y^2) = 0
\]

### Step 7: Identify the surface
This equation represents a **hyperboloid of two sheets**, which is a type of quadratic surface where the distance from the plane and point follows the specified relationship.

### Do you want more details or have any questions?  
Here are five related questions:

1. How would the surface change if the distance ratio was different?
2. What is the geometric interpretation of a hyperboloid?
3. How can you visualize the relationship between points on this surface?
4. How do hyperboloids relate to ellipsoids and paraboloids?
5. What happens to the surface if the plane is shifted?

**Tip:** When solving distance-related problems, always think about using distance formulas and squaring to eliminate roots.

Find an equation of the surface consisting of all points P(x, y, z) that are twice as far from the plane z = -4 as from the point (0, 0, 4). Identify the surface.

Explore the detailed solution for finding the equation of a surface where the distance from any point to the plane z = -4 is twice the distance from the point (0, 0, 4). This problem involves geometry and quadratic surfaces, yielding a hyperboloid of two sheets. Suitable for college-level students studying multivariable calculus.

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