To find the length of the minor axis of the ellipse given the vertices and the foci, follow these steps:
1. Identify the coordinates of the vertices and foci:
- Vertices: [tex]\((-5, 7)\)[/tex] and [tex]\((-5, -3)\)[/tex]
- Foci: [tex]\((-5, 6)\)[/tex] and [tex]\((-5, -2)\)[/tex]
2. Calculate the length of the major axis:
- The major axis length is the distance between the vertices.
- The vertices have the same x-coordinate, so we only need to find the difference in the y-coordinates:
[tex]\[
|7 - (-3)| = |7 + 3| = 10
\][/tex]
- Hence, the length of the major axis is 10 units.
3. Determine the length of the semi-major axis (a):
- The semi-major axis is half the length of the major axis:
[tex]\[
a = \frac{10}{2} = 5
\][/tex]
4. Calculate the distance between the foci:
- The foci also have the same x-coordinate, so we only consider the difference in the y-coordinates:
[tex]\[
|6 - (-2)| = |6 + 2| = 8
\][/tex]
- Thus, the distance between the foci is 8 units.
5. Determine the length of the semi-major axis in terms of the linear distance (2c):
- Since the distance between the foci is equal to [tex]\(2c\)[/tex]:
[tex]\[
c = \frac{8}{2} = 4
\][/tex]
6. Use the relationship [tex]\(a^2 = b^2 + c^2\)[/tex] to find the length of the semi-minor axis (b):
- Substitute the values of [tex]\(a\)[/tex] and [tex]\(c\)[/tex] into the equation:
[tex]\[
a^2 = b^2 + c^2
\][/tex]
- With [tex]\(a = 5\)[/tex] and [tex]\(c = 4\)[/tex]:
[tex]\[
5^2 = b^2 + 4^2
\][/tex]
[tex]\[
25 = b^2 + 16
\][/tex]
- Solving for [tex]\(b^2\)[/tex]:
[tex]\[
b^2 = 25 - 16 = 9
\][/tex]
- Hence:
[tex]\[
b = \sqrt{9} = 3
\][/tex]
7. Calculate the length of the minor axis:
- The minor axis is twice the length of the semi-minor axis:
[tex]\[
\text{Minor axis length} = 2b = 2 \times 3 = 6
\][/tex]
Therefore, the length of the minor axis of the ellipse is 6 units.
