Show that the equation `y^(2) - 8y - x + 19 = 0 ` represents a parabola . Find its vertex, focus and directrix.

Answer

To show that the equation \( y^2 - 8y - x + 19 = 0 \) represents a parabola and to find its vertex, focus, and directrix, we can follow these steps:

### Step 1: Rearranging the Equation
Start by rearranging the given equation to isolate \( x \):
\[ y^2 - 8y + 19 = x \]

### Step 2: Completing the Square
Next, we will complete the square for the \( y \) terms. The expression \( y^2 - 8y \) can be rewritten as:
\[ y^2 - 8y = (y - 4)^2 - 16 \]
Thus, we can substitute this back into the equation:
\[ (y - 4)^2 - 16 + 19 = x \]
This simplifies to:
\[ (y - 4)^2 + 3 = x \]
Or, rearranging gives:
\[ (y - 4)^2 = x - 3 \]

### Step 3: Identifying the Parabola
The equation \( (y - 4)^2 = x - 3 \) is in the standard form of a parabola that opens to the right, which is given by \( (y - k)^2 = 4p(x - h) \), where \( (h, k) \) is the vertex.

### Step 4: Finding the Vertex
From the equation \( (y - 4)^2 = x - 3 \), we can identify:
- Vertex \( (h, k) = (3, 4) \)

### Step 5: Finding the Focus and Directrix
In the standard form \( (y - k)^2 = 4p(x - h) \), we have \( 4p = 1 \), which gives:
\[ p = \frac{1}{4} \]
The focus of the parabola is located at \( (h + p, k) \):
- Focus: \( (3 + \frac{1}{4}, 4) = \left(\frac{13}{4}, 4\right) \)

The equation of the directrix is given by \( x = h - p \):
- Directrix: \( x = 3 - \frac{1}{4} = \frac{11}{4} \)

### Summary of Results
- The vertex of the parabola is \( (3, 4) \).
- The focus of the parabola is \( \left(\frac{13}{4}, 4\right) \).
- The equation of the directrix is \( x = \frac{11}{4} \).