The question appears to describe the equation of a hyperbola with the given information:

1. Vertices: $(0, -1)$ and $(12, -1)$
2. Focus: One focus is given as $(6 + \sqrt{11}, -1)$

Step-by-Step Solution:

1. Determine the center:

The center of the hyperbola lies midway between the two vertices: $\text{Center} = \left( \frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2} \right) = \left( \frac{0 + 12}{2}, \frac{-1 + (-1)}{2} \right) = (6, -1)$

2. Determine the distance between vertices (2a):

The distance between the vertices is the length of the transverse axis, which equals $2a$: $2a = |x_2 - x_1| = |12 - 0| = 12 \quad \Rightarrow \quad a = 6$

3. Determine the distance to the focus (c):

The focus is located at $(6 + \sqrt{11}, -1)$. The distance from the center to the focus is: $c = \sqrt{11}$

4. Find $b$ using the relationship $c^2 = a^2 + b^2$:

For a hyperbola, the relationship between $a$, $b$, and $c$ is: $c^2 = a^2 + b^2$ Substituting $c = \sqrt{11}$ and $a = 6$: $(\sqrt{11})^2 = 6^2 + b^2 \quad \Rightarrow \quad 11 = 36 + b^2 \quad \Rightarrow \quad b^2 = 11 - 36 = -25$ Since $b^2$ must be positive, this suggests the problem is misstated. Could you confirm?

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