Classify the following conic sections. Find its center, vertices, foci, directrix (if possible), asymptotes (if possible). Then, sketch its graph. −y2 − 8x + 2y + 9 = 0

We are given the equation:

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

Let's go step by step to classify the conic section and find its center, vertices, foci, directrix (if possible), and asymptotes (if possible).

### Step 1: Rearrange the equation

Rearrange the terms so that all terms involving \( y \) are on one side:

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

Factor out the negative sign from the left-hand side:

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

### Step 2: Complete the square

Now, let's complete the square for the \( y \)-terms. Take half of the coefficient of \( y \), square it, and add/subtract it inside the parentheses.

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

Substitute this into the equation:

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

Distribute the negative sign:

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

Move the constants to the other side:

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

Finally, solve for \( x \):

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

### Step 3: Classify the conic section

This equation is in the form of a **parabola** that opens horizontally. A horizontally oriented parabola has the standard form:

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

where \( (h, k) \) is the vertex, and \( a \) determines the direction and "width" of the parabola. In our case, \( a = -\frac{1}{8} \), \( h = \frac{5}{4} \), and \( k = 1 \).

Thus, we have a parabola that opens to the left because \( a < 0 \).

### Step 4: Key features of the parabola

- **Vertex**: The vertex is \( \left( \frac{5}{4}, 1 \right) \).
- **Focus**: The focus is calculated using 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 is the vertical line given by \( x = h - \frac{1}{4a} \). Therefore:

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

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

### Step 5: Graph the parabola

Since I can't generate the graph right now, here’s how you can sketch it:
- Mark the **vertex** at \( \left( \frac{5}{4}, 1 \right) \).
- Plot the **focus** at \( \left( -\frac{3}{4}, 1 \right) \).
- Draw the **directrix** as a vertical line at \( x = \frac{13}{4} \).
- The parabola opens to the left and is symmetric about the line \( y = 1 \).

---

Do you want more details or have any questions? Here are five questions to explore this further:

1. How would the graph change if the parabola opened upward or downward?
2. What is the role of the directrix and focus in the definition of a parabola?
3. How can you find the equation of the parabola if you are given the vertex and focus?
4. What happens to the parabola if the coefficient \( a \) changes to a positive value?
5. How can you apply these steps to different conic sections like ellipses or hyperbolas?

**Tip:** The distance between the vertex and the focus is the same as the distance between the vertex and the directrix in any parabola.

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

Learn how to classify and solve the parabola equation −y^2 − 8x + 2y + 9 = 0 by completing the square. This problem walks through the process of finding the vertex, focus, and directrix, and provides a guide for sketching the graph. Suitable for high school students in Grades 10-12.

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