Find the locus of a point which moves so that its distance from (1,2,3) is four times its distance from YZ plane

Answer

To find the locus of a point \( P(x, y, z) \) that moves such that its distance from the point \( (1, 2, 3) \) is four times its distance from the YZ-plane, we can follow these steps:

### Step 1: Understand the distances involved
1. The distance from the point \( P(x, y, z) \) to the point \( (1, 2, 3) \) can be calculated using the distance formula:
\[ d = \sqrt{(x - 1)^2 + (y - 2)^2 + (z - 3)^2} \]
2. The distance from the point \( P(x, y, z) \) to the YZ-plane is simply the x-coordinate of the point, which is \( |x| \).

### Step 2: Set up the equation based on the problem statement
According to the problem, the distance from \( P \) to \( (1, 2, 3) \) is four times the distance from \( P \) to the YZ-plane:
\[ \sqrt{(x - 1)^2 + (y - 2)^2 + (z - 3)^2} = 4|x| \]

### Step 3: Square both sides to eliminate the square root
Squaring both sides gives:
\[ (x - 1)^2 + (y - 2)^2 + (z - 3)^2 = 16x^2 \]

### Step 4: Expand both sides
Expanding the left-hand side:
\[ (x - 1)^2 = x^2 - 2x + 1 \]
\[ (y - 2)^2 = y^2 - 4y + 4 \]
\[ (z - 3)^2 = z^2 - 6z + 9 \]
Thus, we have:
\[ x^2 - 2x + 1 + y^2 - 4y + 4 + z^2 - 6z + 9 = 16x^2 \]

### Step 5: Combine like terms
Combining all terms on the left side:
\[ x^2 + y^2 + z^2 - 2x - 4y - 6z + 14 = 16x^2 \]
Rearranging gives:
\[ x^2 - 16x^2 + y^2 - 2x - 4y - 6z + 14 = 0 \]
This simplifies to:
\[ -15x^2 + y^2 - 2x - 4y - 6z + 14 = 0 \]

### Step 6: Rearranging the equation
Rearranging gives:
\[ 15x^2 - 2x + y^2 - 4y - 6z + 14 = 0 \]

### Step 7: Identify the locus
This equation represents a quadric surface in three-dimensional space, which is the locus of the point \( P(x, y, z) \).

### Final Answer
The locus of the point \( P \) is given by the equation:
\[ 15x^2 - 2x + y^2 - 4y - 6z + 14 = 0 \]