Find the vertex, focus, and directrix of the following parabolas:
`y^(2)-2y+8x-23=0`

Answer

To find the vertex, focus, and directrix of the parabola given by the equation \( y^2 - 2y + 8x - 23 = 0 \), we will follow these steps:

### Step 1: Rearranging the Equation
We start with the equation:
\[ y^2 - 2y + 8x - 23 = 0 \]

We can rearrange it to isolate the \( y \) terms:
\[ y^2 - 2y = -8x + 23 \]

### Step 2: Completing the Square
Next, we will complete the square for the \( y \) terms. To do this, we take the coefficient of \( y \) (which is -2), halve it to get -1, and then square it to get 1. We add and subtract 1 on the left side:
\[ y^2 - 2y + 1 - 1 = -8x + 23 \]
This simplifies to:
\[ (y - 1)^2 - 1 = -8x + 23 \]
Now, we can rewrite it as:
\[ (y - 1)^2 = -8x + 24 \]
Or:
\[ (y - 1)^2 = -8(x - 3) \]

### Step 3: Identifying the Vertex
From the equation \( (y - 1)^2 = -8(x - 3) \), we can identify the vertex of the parabola. The vertex form of a parabola is given by \( (y - k)^2 = 4p(x - h) \), where \( (h, k) \) is the vertex.

Here, we have:
- \( h = 3 \)
- \( k = 1 \)

Thus, the vertex is:
\[ \text{Vertex} = (3, 1) \]

### Step 4: Finding the Value of \( a \)
In the equation \( (y - 1)^2 = -8(x - 3) \), we can see that \( 4p = -8 \). Therefore, we can find \( p \):
\[ p = \frac{-8}{4} = -2 \]

### Step 5: Finding the Focus
The focus of the parabola is located at a distance \( p \) from the vertex along the axis of symmetry. Since the parabola opens to the left (indicated by the negative sign), we subtract \( p \) from the x-coordinate of the vertex:
\[ \text{Focus} = (h + p, k) = (3 - 2, 1) = (1, 1) \]

### Step 6: Finding the Directrix
The directrix is a vertical line located at a distance \( p \) from the vertex in the opposite direction of the focus. Thus, we add \( p \) to the x-coordinate of the vertex:
\[ \text{Directrix} = x = h - p = 3 + 2 = 5 \]

### Summary of Results
- **Vertex**: \( (3, 1) \)
- **Focus**: \( (1, 1) \)
- **Directrix**: \( x = 5 \)