Question

Find the vertices of the ellipse centered at the origin with eccentricity e=1/2
and foci at (0,−3)
and (0,3).

Answer 1

To find the vertices of the ellipse given the eccentricity \( e = \frac{1}{2} \) and the foci at \( (0, -3) \) and \( (0, 3) \), we can proceed as follows:

1. **Identify the orientation and center**:
- Since the foci are located at \( (0, -3) \) and \( (0, 3) \), the ellipse is vertical and centered at the origin \( (0, 0) \).

2. **Locate the foci**:
- The distance from the center to each focus is denoted as \( c \). In this case, the distance \( c \) is 3 (the distance from the origin to either focus, \( (0, 3) \) or \( (0, -3) \)).
\[
c = 3
\]

3. **Use the relationship between \( a \), \( b \), and \( c \)**:
- For an ellipse, the relationship between the semi-major axis \( a \), the semi-minor axis \( b \), and the distance to the foci \( c \) is given by:
\[
c = \sqrt{a^2 - b^2}
\]
Since \( e = \frac{c}{a} \), we can also express \( a \) in terms of \( e \) and \( c \):
\[
e = \frac{c}{a} \implies a = \frac{c}{e}
\]

4. **Calculate \( a \)**:
- Substituting the values for \( c \) and \( e \):
\[
a = \frac{3}{\frac{1}{2}} = 6
\]

5. **Calculate \( b \)**:
- Now, use the relationship \( c = \sqrt{a^2 - b^2} \):
\[
3 = \sqrt{6^2 - b^2} \implies 3 = \sqrt{36 - b^2}
\]
Squaring both sides:
\[
9 = 36 - b^2
\]
Rearranging gives:
\[
b^2 = 36 - 9 = 27 \implies b = \sqrt{27} = 3\sqrt{3}
\]

6. **Locate the vertices**:
- The vertices of a vertical ellipse are located at \( (0, \pm a) \). Thus, the vertices are:
\[
(0, 6) \quad \text{and} \quad (0, -6)
\]

In conclusion, the vertices of the ellipse are \( (0, 6) \) and \( (0, -6) \).