Find the vertex, focus, equation of directrix and axis, of parabolas
`y^(2) `- x + 4y + 5 = 0

Answer

To solve the problem of finding the vertex, focus, equation of the directrix, and axis of the parabola given by the equation \( y^2 - x + 4y + 5 = 0 \), we will follow these steps:

### Step 1: Rearrange the equation
We start with the equation:
\[ y^2 - x + 4y + 5 = 0 \]
Rearranging gives:
\[ y^2 + 4y + 5 = x \]

### Step 2: Complete the square for the \(y\) terms
To complete the square for \(y^2 + 4y\):
\[ y^2 + 4y = (y + 2)^2 - 4 \]
Thus, we can rewrite the equation as:
\[ (y + 2)^2 - 4 + 5 = x \implies (y + 2)^2 = x - 1 \]

### Step 3: Identify the standard form of the parabola
The equation \((y + 2)^2 = x - 1\) can be compared to the standard form of a parabola:
\[ (y - k)^2 = 4a(x - h) \]
Here, \(h = 1\), \(k = -2\), and \(4a = 1\).

### Step 4: Find the vertex
The vertex \((h, k)\) is:
\[ \text{Vertex} = (1, -2) \]

### Step 5: Find the value of \(a\)
From \(4a = 1\), we find:
\[ a = \frac{1}{4} \]

### Step 6: Find the focus
The focus of the parabola is located at \((h + a, k)\):
\[ \text{Focus} = \left(1 + \frac{1}{4}, -2\right) = \left(\frac{5}{4}, -2\right) \]

### Step 7: Find the equation of the directrix
The equation of the directrix is given by:
\[ x = h - a \]
Calculating gives:
\[ \text{Directrix} = 1 - \frac{1}{4} = \frac{3}{4} \]

### Step 8: Find the axis of the parabola
The axis of the parabola is the line that passes through the vertex and is parallel to the axis of symmetry. Since this is a horizontal parabola, the axis is:
\[ y = -2 \]

### Summary of Results
- **Vertex**: \((1, -2)\)
- **Focus**: \(\left(\frac{5}{4}, -2\right)\)
- **Directrix**: \(x = \frac{3}{4}\)
- **Axis**: \(y = -2\)