Given the equation of the parabola (y-1)^2  = 32(x+6).  What is the focus?

The equation of the parabola is \((y - 1)^2 = 32(x + 6)\).

This equation is in the standard form for a horizontally oriented parabola:
\[
(y - k)^2 = 4p(x - h)
\]
where \((h, k)\) is the vertex and \(p\) is the distance from the vertex to the focus.

1. **Identify the vertex**: From the equation, the vertex \((h, k)\) is \((-6, 1)\).

2. **Determine \(p\)**: In the equation \((y - k)^2 = 4p(x - h)\), the value of \(4p = 32\). Therefore,
   \[
   p = \frac{32}{4} = 8
   \]
   Since the parabola opens to the right (positive \(p\)), the focus will be 8 units to the right of the vertex.

3. **Find the focus**: The focus is located at \((h + p, k)\). Substituting the values:
   \[
   \text{Focus} = (-6 + 8, 1) = (2, 1)
   \]

Thus, the focus of the parabola is \((2, 1)\).

Would you like more details on any part of this solution? Here are some related questions you might find helpful:

1. What is the directrix of this parabola?
2. How do you derive the standard form of a parabola equation?
3. What are the properties of a parabola with a horizontal orientation?
4. How do you graph a parabola given its equation?
5. How can you find the length of the latus rectum of a parabola?

**Tip**: For horizontal parabolas, the focus and directrix are always to the left or right of the vertex, depending on the direction of the parabola's opening.

Discover how to find the focus of the parabola (y-1)^2 = 32(x+6) using the vertex form of a parabola equation. This problem explores the properties of parabolas and involves calculating the focus based on given parameters. Suitable for advanced high school students.

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