Given the information, it appears we're dealing with an ellipse. Let's break down the details:

1. Vertices: The vertices are at $(-2, -2)$ and $(-2, 8)$. Since the $x$-coordinates of the vertices are the same, the major axis of the ellipse is vertical.

   • The center of the ellipse is the midpoint of the vertices: $\text{Center} = \left( \frac{-2 + (-2)}{2}, \frac{-2 + 8}{2} \right) = (-2, 3)$

2. Length of the major axis: The distance between the two vertices is the length of the major axis, which is $8 - (-2) = 10$. Therefore, the length of the major axis is 10 units.

   • The semi-major axis $a$ is half of this length: $a = \frac{10}{2} = 5$

3. Length of the minor axis: The problem states that the minor axis is 6 units long.

   • The semi-minor axis $b$ is half of this length: $b = \frac{6}{2} = 3$

4. Equation of the ellipse: The general form of the equation of an ellipse centered at $(h, k)$ with a vertical major axis is: $\frac{(x - h)^2}{b^2} + \frac{(y - k)^2}{a^2} = 1$ Plugging in the values: $\frac{(x + 2)^2}{3^2} + \frac{(y - 3)^2}{5^2} = 1$ Simplifying: $\frac{(x + 2)^2}{9} + \frac{(y - 3)^2}{25} = 1$

This is the equation of the ellipse. The graph will show an ellipse centered at $(-2, 3)$, with a vertical major axis of length 10 and a horizontal minor axis of length 6.

Would you like more details or have any questions?

Here are 5 related questions to consider:

1. How would the equation change if the major axis were horizontal instead?
2. What are the foci of the ellipse?
3. How do you calculate the eccentricity of the ellipse?
4. What is the area enclosed by this ellipse?
5. How does changing the length of the minor axis affect the shape of the ellipse?

Tip: To find the foci of the ellipse, use the relationship $c = \sqrt{a^2 - b^2}$, where $c$ is the distance from the center to each focus.