Find the vertex, focus and directrix of the parabola `x^(2)=2(2x+y)`.

Explanation

To find the vertex, focus, and directrix of the parabola given by the equation \( x^2 = 2(2x + y) \), we will follow these steps:

### Step 1: Rewrite the equation
Start by distributing the right side of the equation:
\[ x^2 = 4x + 2y \]

### Step 2: Rearrange the equation
Move all terms to one side to set the equation to zero:
\[ x^2 - 4x - 2y = 0 \]

### Step 3: Complete the square
To complete the square for the \( x \) terms, we take the coefficient of \( x \) (which is -4), divide it by 2 to get -2, and then square it to get 4. We add and subtract 4:
\[ (x^2 - 4x + 4) - 4 - 2y = 0 \]
This simplifies to:
\[ (x - 2)^2 - 4 - 2y = 0 \]
Rearranging gives:
\[ (x - 2)^2 = 2y + 4 \]
Further simplifying leads to:
\[ (x - 2)^2 = 2(y + 2) \]

### Step 4: Identify the vertex
The equation is now in the standard form of a parabola \((x - h)^2 = 4p(y - k)\), where \((h, k)\) is the vertex. From \((x - 2)^2 = 2(y + 2)\), we can see:
- \(h = 2\)
- \(k = -2\)

Thus, the vertex is:
\[ \text{Vertex} = (2, -2) \]

### Step 5: Find the focus
In the standard form, \(4p\) is equal to 2, so:
\[ 4p = 2 \implies p = \frac{2}{4} = \frac{1}{2} \]
Since the parabola opens upwards, the focus will be \(p\) units above the vertex. Therefore, the coordinates of the focus are:
\[ \text{Focus} = \left(2, -2 + \frac{1}{2}\right) = \left(2, -\frac{3}{2}\right) \]

### Step 6: Find the directrix
The directrix is a horizontal line located \(p\) units below the vertex. Therefore, the equation of the directrix is:
\[ y = -2 - \frac{1}{2} = -\frac{5}{2} \]

### Summary of Results
- **Vertex**: \((2, -2)\)
- **Focus**: \((2, -\frac{3}{2})\)
- **Directrix**: \(y = -\frac{5}{2}\)