In this problem, we are dealing with an ellipse. Let's break down the given information to find the equation of the ellipse.

Key points:

• Co-vertices: (7, -3) and (-1, -3)
• Vertex: (3, 3)

Step 1: Determine the center of the ellipse

The center of the ellipse lies at the midpoint of the co-vertices. The midpoint formula is:

$\text{Midpoint} = \left( \frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2} \right)$

Using the coordinates of the co-vertices (7, -3) and (-1, -3):

$\text{Center} = \left( \frac{7 + (-1)}{2}, \frac{-3 + (-3)}{2} \right) = \left( \frac{6}{2}, \frac{-6}{2} \right) = (3, -3)$

So, the center of the ellipse is (3, -3).

Step 2: Determine the orientation of the ellipse

• Since the co-vertices lie along the horizontal line $y = -3$, the minor axis is horizontal.
• The major axis is vertical because the vertex (3, 3) is directly above the center (3, -3).

Step 3: Find the semi-major and semi-minor axes

• The distance from the center (3, -3) to the vertex (3, 3) is the semi-major axis $a$. This distance is:

$a = |3 - (-3)| = 6$

• The distance between the co-vertices (7, -3) and (-1, -3) gives the length of the entire minor axis. The semi-minor axis $b$ is half of that distance:

$b = \frac{7 - (-1)}{2} = \frac{8}{2} = 4$

Step 4: Write the equation of the ellipse

The equation of an ellipse centered at $(h, k)$ with a vertical major axis is:

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

Substituting $h = 3$, $k = -3$, $a = 6$, and $b = 4$:

$\frac{(x - 3)^2}{4^2} + \frac{(y + 3)^2}{6^2} = 1$

Simplifying:

$\frac{(x - 3)^2}{16} + \frac{(y + 3)^2}{36} = 1$

Thus, the equation of the ellipse is:

$\boxed{\frac{(x - 3)^2}{16} + \frac{(y + 3)^2}{36} = 1}$

Would you like more details on how ellipses are derived or any other specific questions?

Related Questions:

1. How do you determine the foci of an ellipse?
2. What is the eccentricity of the ellipse in this problem?
3. How would the equation change if the major axis was horizontal?
4. How can you graph this ellipse based on the equation?
5. How does the standard form of an ellipse compare to the form of a circle?

Tip: The larger denominator in the equation of an ellipse always corresponds to the square of the semi-major axis, indicating its orientation.