Find the vertex, focus, latus-rectum axis and directrix of the parabola `x^(2) - y - 2x = 0.`

Answer

To find the vertex, focus, latus rectum, axis, and directrix of the parabola given by the equation \(x^2 - y - 2x = 0\), we will follow these steps:

### Step 1: Rearranging the Equation
First, we rearrange the equation to express \(y\) in terms of \(x\):
\[ x^2 - 2x - y = 0 \implies y = x^2 - 2x \]

### Step 2: Completing the Square
Next, we complete the square for the expression on the right side:
\[ y = (x^2 - 2x + 1) - 1 = (x - 1)^2 - 1 \]
This can be rewritten as:
\[ y + 1 = (x - 1)^2 \]

### Step 3: Identifying the Standard Form
Now, we can identify the standard form of the parabola:
\[ (y - k) = a(x - h)^2 \]
From our equation, we have:
- \(h = 1\)
- \(k = -1\)
- \(a = 1\)

### Step 4: Finding the Vertex
The vertex \((h, k)\) of the parabola is:
\[ \text{Vertex} = (1, -1) \]

### Step 5: Finding the Focus
For a parabola of the form \((y - k) = a(x - h)^2\), the focus is located at:
\[ \left(h, k + \frac{1}{4a}\right) \]
Here, \(a = 1\), so:
\[ \text{Focus} = \left(1, -1 + \frac{1}{4}\right) = \left(1, -\frac{3}{4}\right) \]

### Step 6: Finding the Directrix
The equation of the directrix for this parabola is given by:
\[ y = k - \frac{1}{4a} \]
Thus:
\[ \text{Directrix} = y = -1 - \frac{1}{4} = -\frac{5}{4} \]

### Step 7: Finding the Axis of Symmetry
The axis of symmetry for a parabola of this form is the vertical line:
\[ x = h = 1 \]

### Step 8: Finding the Length of the Latus Rectum
The length of the latus rectum is given by:
\[ \text{Length of Latus Rectum} = \frac{1}{a} = 4 \]

### Summary of Results
- **Vertex**: \((1, -1)\)
- **Focus**: \((1, -\frac{3}{4})\)
- **Directrix**: \(y = -\frac{5}{4}\)
- **Axis of Symmetry**: \(x = 1\)
- **Length of Latus Rectum**: \(4\)