Transform to parallel axes through the point (1, -2) the equations
`y^(2) + 4x + 4y + 8 = 0`

Answer

To transform the equation \( y^2 + 4x + 4y + 8 = 0 \) to parallel axes through the point \( (1, -2) \), we will follow these steps:

### Step 1: Define the transformation
We need to shift the origin to the point \( (1, -2) \). Let the new coordinates be \( (x', y') \). The relationships between the old coordinates \( (x, y) \) and the new coordinates \( (x', y') \) are given by:
\[ x = x' + 1 \]
\[ y = y' - 2 \]

### Step 2: Substitute the transformations into the original equation
We will substitute \( x \) and \( y \) in the original equation \( y^2 + 4x + 4y + 8 = 0 \):
\[ y^2 = (y' - 2)^2 = y'^2 - 4y' + 4 \]
\[ x = x' + 1 \]
\[ y = y' - 2 \]

Substituting these into the equation:
\[ (y' - 2)^2 + 4(x' + 1) + 4(y' - 2) + 8 = 0 \]

### Step 3: Expand the equation
Now, we will expand the equation:
\[ y'^2 - 4y' + 4 + 4(x' + 1) + 4y' - 8 + 8 = 0 \]
This simplifies to:
\[ y'^2 - 4y' + 4 + 4x' + 4 + 4y' - 8 + 8 = 0 \]

### Step 4: Combine like terms
Combining the like terms:
\[ y'^2 + (4y' - 4y') + (4x' + 4 + 4 - 8 + 8) = 0 \]
This simplifies to:
\[ y'^2 + 4x' + 8 = 0 \]

### Step 5: Final equation
Thus, the transformed equation is:
\[ y'^2 + 4x' + 8 = 0 \]