The problem involves finding the equation of a hyperbola with given vertices and foci. Let's solve it step by step:

Given Information:

1. Vertices:

   • $V_1(-1, -1)$
   • $V_2(5, -1)$

2. Foci:

   • $F_1(-4, -1)$
   • $F_2(8, -1)$

Step 1: Determine the orientation

• Since both the vertices and foci share the same $y$-coordinate ($-1$), the hyperbola is horizontal.

The general equation for a horizontal hyperbola is: $\frac{(x-h)^2}{a^2} - \frac{(y-k)^2}{b^2} = 1$

Here, $(h, k)$ is the center, $a$ is the distance from the center to a vertex, and $c$ is the distance from the center to a focus.

Step 2: Calculate the center

The center, $(h, k)$, is the midpoint of the vertices: $h = \frac{-1 + 5}{2} = 2, \quad k = \frac{-1 + (-1)}{2} = -1$ So, the center is: $(2, -1)$

Step 3: Calculate $a$

The distance between the center and either vertex is $a$: $a = \text{distance from center to vertex} = |2 - (-1)| = 3$ Thus, $a = 3$.

Step 4: Calculate $c$

The distance from the center to a focus is $c$: $c = \text{distance from center to focus} = |2 - (-4)| = 6$ Thus, $c = 6$.

Step 5: Calculate $b$

Using the relationship $c^2 = a^2 + b^2$ for hyperbolas: $c^2 = a^2 + b^2 \implies 6^2 = 3^2 + b^2 \implies 36 = 9 + b^2 \implies b^2 = 27$

Step 6: Write the equation

Now substitute $h = 2$, $k = -1$, $a^2 = 9$, and $b^2 = 27$ into the standard form: $\frac{(x-2)^2}{9} - \frac{(y+1)^2}{27} = 1$

Final Answer:

The equation of the hyperbola is: $\boxed{\frac{(x-2)^2}{9} - \frac{(y+1)^2}{27} = 1}$

Let me know if you'd like further details or have any questions! 😊

Related Questions:

1. How is the orientation of a hyperbola determined?
2. What is the relationship between $a$, $b$, and $c$ for a hyperbola?
3. How can you find the center of a hyperbola given the vertices?
4. What happens to the equation if the hyperbola is vertical instead of horizontal?
5. How do you verify if an equation represents a hyperbola?

Tip:

Always start by identifying the orientation (horizontal or vertical) based on the given coordinates of the vertices and foci. This simplifies the process significantly!