We are tasked with finding the equation of a hyperbola with the following properties:

1. Vertices: $(-2, -4)$ and $(12, -4)$
2. Asymptote: $y + 4 = \frac{8}{7}(x - 5)$

Step 1: Identify the center

The center $(h, k)$ is the midpoint of the vertices. Using the midpoint formula:

$h = \frac{-2 + 12}{2}, \quad k = \frac{-4 + (-4)}{2}$

$h = 5, \quad k = -4$

So, the center is $(5, -4)$.

Step 2: Identify the orientation

Since the vertices have the same $y$-coordinate ($k = -4$), the transverse axis is horizontal. This means the equation of the hyperbola will have the form:

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

Step 3: Find $a^2$

The distance between the vertices is $2a$, so:

$2a = 12 - (-2) = 14 \quad \Rightarrow \quad a = \frac{14}{2} = 7$

Thus, $a^2 = 7^2 = 49$.

Step 4: Use the asymptote to find $b^2$

The slope of the asymptote is given as $\frac{8}{7}$. For a hyperbola with a horizontal transverse axis, the slope of the asymptotes is given by:

$\text{slope} = \pm \frac{b}{a}$

Substituting the known slope:

$\frac{b}{a} = \frac{8}{7} \quad \Rightarrow \quad b = \frac{8}{7} \cdot 7 = 8$

Thus, $b^2 = 8^2 = 64$.

Step 5: Write the equation

Substitute the values of $h = 5$, $k = -4$, $a^2 = 49$, and $b^2 = 64$ into the general form:

$\frac{(x-5)^2}{49} - \frac{(y+4)^2}{64} = 1$

This is the equation of the hyperbola.

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Let me know if you'd like me to assist in graphing this hyperbola or further clarifications!

Additional questions:

1. Can you identify the foci of this hyperbola?
2. How would the equation change if the transverse axis were vertical instead?
3. What are the equations of the other asymptote?
4. How does the slope of the asymptote affect the equation of the hyperbola?
5. How would you verify if a given point lies on this hyperbola?

Tip: Always verify hyperbola equations by checking if the slopes of the asymptotes match the provided information.