https://file.aiquickdraw.com/mathbot/file/1727432479002f8tyyoza.jpg

The equation in the image represents a **hyperbola** in its standard form. Here's the equation:

\[
\frac{(x - 4)^2}{9} - \frac{(y + 5)^2}{16} = 1
\]

This is the equation of a hyperbola centered at \( (4, -5) \), and since the \(x\)-term is positive, it opens horizontally.

### Key components:
1. **Center**: The center of the hyperbola is \( (h, k) = (4, -5) \).
2. **\(a^2\)** and **\(b^2\)**: The denominators represent \(a^2\) and \(b^2\):
   - \(a^2 = 9\), so \(a = 3\), representing the distance from the center to the vertices along the \(x\)-axis.
   - \(b^2 = 16\), so \(b = 4\), representing the distance from the center to the co-vertices along the \(y\)-axis.
3. **Vertices**: The vertices are located at \( (4 - 3, -5) = (1, -5) \) and \( (4 + 3, -5) = (7, -5) \).
4. **Asymptotes**: The equations of the asymptotes for this hyperbola are given by:
   \[
   y + 5 = \pm \frac{b}{a}(x - 4)
   \]
   Substituting \(a = 3\) and \(b = 4\):
   \[
   y + 5 = \pm \frac{4}{3}(x - 4)
   \]

Would you like further details or explanations?

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Here are 5 related questions:
1. How do you find the foci of the hyperbola?
2. What is the difference between a hyperbola and an ellipse?
3. How can the asymptotes help graph a hyperbola?
4. How does changing the values of \(a\) and \(b\) affect the shape of the hyperbola?
5. What are real-life applications of hyperbolas?

**Tip**: The value of \(c\) (the distance from the center to the foci) is found using the formula \(c = \sqrt{a^2 + b^2}\).

The query submitted by the user: Solve the hyperbola equation \(\frac{(x - 4)^2}{9} - \frac{(y + 5)^2}{16} = 1\)

Learn how to solve and analyze the hyperbola equation \(\frac{(x - 4)^2}{9} - \frac{(y + 5)^2}{16} = 1\). This guide covers key mathematical concepts like finding the center, vertices, asymptotes, and foci of the hyperbola. Suitable for students in Grades 10-12.

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