https://file.aiquickdraw.com/mathbot/file/1727433200751qdjnq0e8.jpg

The equation shown in the image is:

\[
\frac{(y - 9)^2}{24} - \frac{(x + 5)^2}{8} = 1
\]

This represents the equation of a hyperbola. Here's how we can interpret it:

### General form of a hyperbola:
\[
\frac{(y - k)^2}{a^2} - \frac{(x - h)^2}{b^2} = 1
\]
This is the equation of a vertically oriented hyperbola centered at \((h, k)\). 

From the given equation:
- \((x + 5)\) suggests \(h = -5\)
- \((y - 9)\) suggests \(k = 9\)
- \(a^2 = 24\), so \(a = \sqrt{24} = 2\sqrt{6}\)
- \(b^2 = 8\), so \(b = \sqrt{8} = 2\sqrt{2}\)

### Characteristics:
- **Center**: \((-5, 9)\)
- **Vertices**: Since it's vertically oriented, the vertices are \(a\) units above and below the center. Thus, the vertices are at \((-5, 9 + 2\sqrt{6})\) and \((-5, 9 - 2\sqrt{6})\).
- **Asymptotes**: The equations of the asymptotes are given by:
  \[
  y - 9 = \pm \frac{a}{b}(x + 5)
  \]
  Substituting \(a = 2\sqrt{6}\) and \(b = 2\sqrt{2}\), the asymptotes become:
  \[
  y - 9 = \pm \sqrt{3}(x + 5)
  \]

Would you like further details, or do you have any questions?

Here are 5 related questions:
1. How can we find the foci of this hyperbola?
2. What is the eccentricity of the hyperbola?
3. How do asymptotes help in sketching a hyperbola?
4. How does the orientation of a hyperbola change based on the sign of its terms?
5. Can you derive the length of the transverse and conjugate axes for this hyperbola?

**Tip:** Hyperbolas are defined by their eccentricity, which is always greater than 1!

Solve the equation: (y - 9)^2 / 24 - (x + 5)^2 / 8 = 1.

This solution explains how to solve the hyperbola equation (y - 9)^2 / 24 - (x + 5)^2 / 8 = 1. Learn key concepts of hyperbolas, vertices, asymptotes, and their properties. Suitable for Grades 11-12 students studying conic sections.

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