Find the locus of a point whose distance from `x`-axis is twice the distance from the point (1, -1, 2)

Answer

To find the locus of a point whose distance from the x-axis is twice the distance from the point (1, -1, 2), we can follow these steps:

### Step 1: Define the point and distances
Let the point be \( P(a, b, c) \). The distance from the x-axis (denoted as \( l_1 \)) is given by the formula:
\[ l_1 = \sqrt{b^2 + c^2} \]
This is because the distance from the x-axis is determined by the y-coordinate (b) and the z-coordinate (c).

### Step 2: Calculate the distance from the point (1, -1, 2)
The distance from the point \( P(a, b, c) \) to the point \( (1, -1, 2) \) (denoted as \( l_2 \)) is given by:
\[ l_2 = \sqrt{(a - 1)^2 + (b + 1)^2 + (c - 2)^2} \]

### Step 3: Set up the relationship between the distances
According to the problem, the distance from the x-axis is twice the distance from the point (1, -1, 2):
\[ l_1 = 2l_2 \]
Substituting the expressions for \( l_1 \) and \( l_2 \):
\[ \sqrt{b^2 + c^2} = 2\sqrt{(a - 1)^2 + (b + 1)^2 + (c - 2)^2} \]

### Step 4: Square both sides to eliminate the square roots
Squaring both sides gives:
\[ b^2 + c^2 = 4((a - 1)^2 + (b + 1)^2 + (c - 2)^2) \]

### Step 5: Expand the right-hand side
Expanding the right-hand side:
\[ b^2 + c^2 = 4[(a - 1)^2 + (b + 1)^2 + (c - 2)^2] \]
\[ = 4[(a^2 - 2a + 1) + (b^2 + 2b + 1) + (c^2 - 4c + 4)] \]
\[ = 4[a^2 - 2a + b^2 + 2b + c^2 + 5] \]
\[ = 4a^2 - 8a + 4b^2 + 8b + 4c^2 + 20 \]

### Step 6: Rearrange the equation
Now, we can rearrange the equation:
\[ b^2 + c^2 = 4a^2 - 8a + 4b^2 + 8b + 4c^2 + 20 \]
Bringing all terms to one side gives:
\[ b^2 + c^2 - 4b^2 - 4c^2 + 8a - 4a^2 - 8b - 20 = 0 \]
\[ -3b^2 - 3c^2 - 4a^2 + 8a - 8b - 20 = 0 \]

### Step 7: Simplify the equation
Dividing through by -1:
\[ 4a^2 + 3b^2 + 3c^2 - 8a + 8b + 20 = 0 \]

### Step 8: Final form of the locus
This can be rearranged to:
\[ 4x^2 + 3y^2 + 3z^2 - 8x + 8y + 20 = 0 \]
This is the equation of the locus of the point.

### Summary of the solution
The locus of the point is given by:
\[ 4x^2 + 3y^2 + 3z^2 - 8x + 8y + 20 = 0 \]