Find the vertex, focus, and directrix of the following parabolas:
`x^(2)+8x+12y+4=0`

Answer

To find the vertex, focus, and directrix of the parabola given by the equation \( x^2 + 8x + 12y + 4 = 0 \), we will follow these steps:

### Step 1: Rearranging the Equation
We start by rearranging the equation to isolate the \( y \) term.
\[ x^2 + 8x + 12y + 4 = 0 \implies 12y = -x^2 - 8x - 4 \implies y = -\frac{1}{12}(x^2 + 8x + 4) \]

### Step 2: Completing the Square
Next, we complete the square for the expression involving \( x \):
\[ x^2 + 8x = (x + 4)^2 - 16 \]
Substituting this back into the equation gives:
\[ y = -\frac{1}{12}((x + 4)^2 - 16 + 4) = -\frac{1}{12}((x + 4)^2 - 12) \]
This simplifies to:
\[ y = -\frac{1}{12}(x + 4)^2 + 1 \]

### Step 3: Identifying the Vertex
From the standard form of the parabola \( y = a(x - h)^2 + k \), we can identify the vertex:
- Here, \( h = -4 \) and \( k = 1 \).
- Therefore, the vertex is at the point \( (-4, 1) \).

### Step 4: Finding the Focus
The standard form of the parabola also allows us to find the focus. The distance \( a \) from the vertex to the focus is given by:
\[ a = \frac{1}{4p} \implies p = -3 \quad (\text{since } a = -\frac{1}{12}) \]
Thus, the focus is located at:
\[ \text{Focus} = (h, k - p) = (-4, 1 - 3) = (-4, -2) \]

### Step 5: Finding the Directrix
The directrix of the parabola can be found using the formula:
\[ y = k + p \implies y = 1 + 3 = 4 \]

### Summary of Results
- **Vertex**: \( (-4, 1) \)
- **Focus**: \( (-4, -2) \)
- **Directrix**: \( y = 4 \)