3. The foci of an ellipse are (-2,-1) and (-2,1). For any point on the ellipse, the sum of its distances from the foci is 8. Determine the orientation, center, vertices, covertices, the lengths of the major and minor axis, and the standard equation of the ellipse.

To analyze the given ellipse, we need to extract and compute the necessary information based on the data provided:

### Step 1: **Determine the orientation and center**
- The foci of the ellipse are given as \((-2, -1)\) and \((-2, 1)\).
- Since the x-coordinates of the foci are the same (\(-2\)), the major axis is vertical. 
- The center of the ellipse is the midpoint of the segment connecting the foci.

\[
\text{Center} = \left(\frac{-2 + (-2)}{2}, \frac{-1 + 1}{2}\right) = (-2, 0)
\]

### Step 2: **Calculate the distance between the foci**
- The distance between the foci is given by \(2c\), where \(c\) is half the distance between the foci:

\[
c = \frac{\text{Distance between the foci}}{2} = \frac{|1 - (-1)|}{2} = \frac{2}{2} = 1
\]

### Step 3: **Calculate the length of the major axis**
- The sum of the distances from any point on the ellipse to the two foci is given as \(2a = 8\). Therefore, \(a\) (the semi-major axis) is:

\[
a = \frac{8}{2} = 4
\]

### Step 4: **Calculate the length of the minor axis**
- To find \(b\) (the semi-minor axis), use the relationship \(b^2 = a^2 - c^2\):

\[
b^2 = 4^2 - 1^2 = 16 - 1 = 15
\]
\[
b = \sqrt{15}
\]

### Step 5: **Determine the vertices and covertices**
- Since the ellipse is vertically oriented:
  - The vertices are at \((h, k \pm a)\):
  
\[
\text{Vertices} = (-2, 0 \pm 4) = (-2, 4) \text{ and } (-2, -4)
\]

  - The covertices are at \((h \pm b, k)\):
  
\[
\text{Covertices} = \left(-2 \pm \sqrt{15}, 0\right) = \left(-2 + \sqrt{15}, 0\right) \text{ and } \left(-2 - \sqrt{15}, 0\right)
\]

### Step 6: **Write the standard equation of the ellipse**
- For a vertical ellipse centered at \((h, k)\), the standard form is:

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

Substituting \(h = -2\), \(k = 0\), \(a = 4\), and \(b = \sqrt{15}\):

\[
\frac{(x + 2)^2}{15} + \frac{y^2}{16} = 1
\]

### Summary of results:

- **Orientation**: Vertical
- **Center**: \((-2, 0)\)
- **Vertices**: \((-2, 4)\) and \((-2, -4)\)
- **Covertices**: \(\left(-2 + \sqrt{15}, 0\right)\) and \(\left(-2 - \sqrt{15}, 0\right)\)
- **Length of Major Axis**: \(2a = 8\)
- **Length of Minor Axis**: \(2b = 2\sqrt{15}\)
- **Standard Equation**: \(\frac{(x + 2)^2}{15} + \frac{y^2}{16} = 1\)

Would you like more details or have any questions?

Here are 5 related questions:
1. How do you determine the foci of an ellipse given its equation?
2. What is the significance of the sum of the distances from any point on the ellipse to the foci?
3. How do you find the eccentricity of an ellipse?
4. Can you derive the equation of an ellipse given the lengths of the major and minor axes?
5. What changes in the equation if the ellipse were horizontally oriented?

**Tip**: When dealing with ellipses, remember that the major axis always corresponds to the larger of the two values \(a\) or \(b\), regardless of orientation.

Explore how to analyze ellipses using given foci and distances, determining orientation, center, vertices, and equation. Learn about ellipse geometry and equations. Suitable for Grades 10-12.

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