To find the equation of the ellipse with the given co-vertices $(7, -3)$ and $(-1, -3)$, and the vertex at $(3, 3)$, we need to follow these steps:

Step 1: Determine the center of the ellipse

The co-vertices lie on the minor axis. The center of the ellipse lies at the midpoint between the two co-vertices.

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

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

Step 2: Determine the semi-major axis $a$

The vertex $(3, 3)$ lies on the major axis, which is vertical (since the y-coordinate changes). The distance between the center and the vertex gives us the length of the semi-major axis $a$:

$a = \text{distance between the center and the vertex} = |3 - (-3)| = 6$

Step 3: Determine the semi-minor axis $b$

The distance between the co-vertices gives us the length of the minor axis. The length of the entire minor axis is the distance between the co-vertices $(7, -3)$ and $(-1, -3)$, which is 8 units. Hence, the semi-minor axis $b$ is half of this:

$b = \frac{8}{2} = 4$

Step 4: Write the equation of the ellipse

Since the major axis is vertical and the center is $(3, -3)$, the general form of the ellipse equation is:

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

Where $(h, k)$ is the center, $a$ is the length of the semi-major axis, and $b$ is the length of the semi-minor axis. Substituting the known values:

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

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

Step 5: Determine the foci

The foci lie along the major axis. The distance from the center to each focus is given by $c$, where $c$ is determined using the formula:

$c^2 = a^2 - b^2$

$c^2 = 6^2 - 4^2 = 36 - 16 = 20$

$c = \sqrt{20} = 2\sqrt{5}$

Since the major axis is vertical, the foci are located at $(3, -3 \pm 2\sqrt{5})$.

Final Answer:

• The equation of the ellipse is: $\frac{(x - 3)^2}{16} + \frac{(y + 3)^2}{36} = 1$
• The foci are at $(3, -3 + 2\sqrt{5})$ and $(3, -3 - 2\sqrt{5})$.

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Related Questions:

1. How do you derive the equation for a horizontal ellipse?
2. What is the geometric significance of the foci in an ellipse?
3. How do you calculate the eccentricity of an ellipse?
4. What happens to an ellipse's equation when its center is moved to the origin?
5. How does the relationship between $a$, $b$, and $c$ affect the shape of an ellipse?

Tip:

The closer the foci are to the center, the more circular the ellipse becomes.