Classify the following conic sections. Find its center, vertices, foci, directrix (if
possible), asymptotes (if possible). Then, sketch its graph.−y
2 − 8x + 2y + 9 = 0.

We are given the equation:

\[
-y^2 - 8x + 2y + 9 = 0
\]

Let's classify the conic section and extract relevant information like the center, vertices, foci, directrix, and asymptotes (if applicable).

### Step 1: Rearranging the equation
We start by rearranging the equation to bring it into a recognizable standard form. First, isolate the \( x \)-related terms on one side:

\[
-y^2 + 2y = 8x - 9
\]

Factor out the negative sign from the left side:

\[
-(y^2 - 2y) = 8x - 9
\]

### Step 2: Completing the square
Now, complete the square for the \( y \)-terms inside the parentheses. To complete the square, take half of the coefficient of \( y \) (which is \( -2 \)), square it, and add/subtract inside the parentheses.

\[
y^2 - 2y \quad \rightarrow \quad \left( y - 1 \right)^2 - 1
\]

So, the equation becomes:

\[
-( (y - 1)^2 - 1) = 8x - 9
\]

Distribute the negative sign:

\[
-(y - 1)^2 + 1 = 8x - 9
\]

Now, move the constant terms to the right-hand side:

\[
-(y - 1)^2 = 8x - 10
\]

Finally, solve for \( x \):

\[
x = -\frac{1}{8}(y - 1)^2 + \frac{5}{4}
\]

### Step 3: Identify the conic
This is the equation of a **parabola** that opens horizontally. The standard form of a horizontal parabola is:

\[
x = a(y - k)^2 + h
\]

where \( (h, k) \) is the vertex, and \( a \) determines the direction and "width" of the parabola.

Here, \( h = \frac{5}{4} \), \( k = 1 \), and \( a = -\frac{1}{8} \).

### Step 4: Key features of the parabola

- **Center (Vertex)**: The vertex of the parabola is \( \left( \frac{5}{4}, 1 \right) \).
- **Focus**: For a parabola in the form \( x = a(y - k)^2 + h \), the focus is given by the formula \( \left( h + \frac{1}{4a}, k \right) \). Here, \( a = -\frac{1}{8} \), so:

\[
\text{Focus} = \left( \frac{5}{4} + \frac{1}{4(-\frac{1}{8})}, 1 \right) = \left( \frac{5}{4} - 2, 1 \right) = \left( -\frac{3}{4}, 1 \right)
\]

- **Directrix**: The directrix of the parabola is the vertical line given by \( x = h - \frac{1}{4a} \). Thus:

\[
\text{Directrix} = x = \frac{5}{4} + 2 = \frac{13}{4}
\]

- **Axis of symmetry**: Since the parabola opens horizontally, its axis of symmetry is the horizontal line \( y = 1 \).

### Step 5: Sketch of the graph

This parabola opens to the left (since \( a = -\frac{1}{8} \)), with the vertex at \( \left( \frac{5}{4}, 1 \right) \), the focus at \( \left( -\frac{3}{4}, 1 \right) \), and the directrix at \( x = \frac{13}{4} \).

I can provide a sketch of this parabola to visualize it clearly. Would you like me to generate that for you?

---

Do you want any further details or have any questions about the solution?

Here are five related questions to further explore this topic:
1. How would the graph change if the parabola opened vertically?
2. Can you derive the equation of a parabola given a different vertex and focus?
3. How is the directrix related to the focus for a parabola?
4. What happens if the coefficient \( a \) is positive instead of negative?
5. How does the method change for conic sections like ellipses or hyperbolas?

**Tip:** For any parabola, the distance from the vertex to the focus is the same as the distance from the vertex to the directrix.

Classify the following conic sections. Find its center, vertices, foci, directrix (if possible), asymptotes (if possible). Then, sketch its graph. −y^2 − 8x + 2y + 9 = 0.

This problem demonstrates how to classify and graph the conic section −y^2 − 8x + 2y + 9 = 0, including finding the center, vertices, focus, directrix, and axis of symmetry. Follow a step-by-step solution involving completing the square and identifying key features of a horizontally opening parabola.

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