https://file.aiquickdraw.com/mathbot/file/1727432411445ekepi6k8.jpg

The given equation is:

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

This is the equation of a hyperbola. Let's analyze it step-by-step.

### General Form of a Hyperbola:
The general form of a hyperbola centered at \((h, k)\) with a horizontal transverse axis is:

\[
\frac{(x - h)^2}{a^2} - \frac{(y - k)^2}{b^2} = 1
\]

Here:
- \( h = 4 \)
- \( k = -5 \)
- \( a^2 = 9 \), so \( a = 3 \)
- \( b^2 = 16 \), so \( b = 4 \)

### Characteristics of the Hyperbola:
- **Center**: \((4, -5)\)
- **Vertices**: The vertices are located \(a\) units from the center along the x-axis. Therefore, the vertices are at:
  \[
  (4 + 3, -5) = (7, -5) \quad \text{and} \quad (4 - 3, -5) = (1, -5)
  \]
- **Foci**: The foci are calculated using the formula \(c^2 = a^2 + b^2\), where \(c\) is the distance from the center to the foci. So,
  \[
  c^2 = 9 + 16 = 25 \quad \Rightarrow \quad c = 5
  \]
  The foci are located \(c\) units from the center along the x-axis. Therefore, the foci are at:
  \[
  (4 + 5, -5) = (9, -5) \quad \text{and} \quad (4 - 5, -5) = (-1, -5)
  \]

- **Asymptotes**: The asymptotes for this hyperbola are lines through the center with slopes \(\pm \frac{b}{a} = \pm \frac{4}{3}\). The equations of the asymptotes are:
  \[
  y + 5 = \pm \frac{4}{3}(x - 4)
  \]
  Simplifying, we get the equations:
  \[
  y = \frac{4}{3}(x - 4) - 5 \quad \text{and} \quad y = -\frac{4}{3}(x - 4) - 5
  \]

### Summary:
- Center: \((4, -5)\)
- Vertices: \((7, -5)\) and \((1, -5)\)
- Foci: \((9, -5)\) and \((-1, -5)\)
- Asymptotes: \(y = \frac{4}{3}(x - 4) - 5\) and \(y = -\frac{4}{3}(x - 4) - 5\)

Let me know if you want more details or have any questions!

Here are 5 related questions to explore:
1. How do you calculate the eccentricity of this hyperbola?
2. What is the significance of the asymptotes in the graph of a hyperbola?
3. Can you plot this hyperbola along with its asymptotes?
4. How would the equation change if the transverse axis were vertical?
5. What is the equation of the conjugate hyperbola for this setup?

**Tip**: Always ensure you determine whether the transverse axis is horizontal or vertical by examining the terms in the hyperbola equation!

Solve the hyperbola equation: (x - 4)^2 / 9 - (y + 5)^2 / 16 = 1.

Learn how to solve the hyperbola equation (x - 4)^2 / 9 - (y + 5)^2 / 16 = 1 with a detailed solution. Key concepts include finding the center, vertices, foci, and asymptotes. Suitable for students in Grades 11-12.

Math Problem Statement

Solution

Ask a new question for Free

By Image

Math Problem Analysis

Mathematical Concepts

Formulas

Theorems

Suitable Grade Level

Related Recommendation