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

Answer

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

### Step 1: Identify the transformation
We need to shift the coordinate system so that the new axes are parallel to the original axes but pass through the point \((1, -2)\). We define the new coordinates as:
\[ x' = x - 1 \quad \text{and} \quad y' = y + 2 \]
This means:
\[ x = x' + 1 \quad \text{and} \quad y = y' - 2 \]

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

### Step 3: Expand the equation
Now, we will expand each term:
1. \(2(x' + 1)^2 = 2(x'^2 + 2x' + 1) = 2x'^2 + 4x' + 2\)
2. \((y' - 2)^2 = y'^2 - 4y' + 4\)
3. \(-4(x' + 1) = -4x' - 4\)
4. \(4(y' - 2) = 4y' - 8\)

Putting it all together:
\[ 2x'^2 + 4x' + 2 + y'^2 - 4y' + 4 - 4x' - 4 + 4y' - 8 = 0 \]

### Step 4: Combine like terms
Now, we combine like terms:
\[ 2x'^2 + y'^2 + (4x' - 4x') + (-4y' + 4y') + (2 + 4 - 4 - 8) = 0 \]
This simplifies to:
\[ 2x'^2 + y'^2 - 6 = 0 \]

### Step 5: Rearranging the equation
We can rearrange the equation to get:
\[ 2x'^2 + y'^2 = 6 \]

### Final Result
Thus, the transformed equation in the new coordinate system is:
\[ 2x'^2 + y'^2 = 6 \]

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