Certainly! Let's solve the problem step-by-step.
### Equation of the Ellipse
The given equation of the ellipse is:
[tex]\[
\frac{x^2}{100} + \frac{y^2}{9} = 1
\][/tex]
This equation is in the standard form of an ellipse centered at the origin, [tex]\(\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1\)[/tex].
In this case:
[tex]\[ a^2 = 100 \][/tex]
[tex]\[ b^2 = 9 \][/tex]
### Finding [tex]\(a\)[/tex] and [tex]\(b\)[/tex]
To find [tex]\(a\)[/tex] and [tex]\(b\)[/tex]:
[tex]\[ a = \sqrt{100} = 10 \][/tex]
[tex]\[ b = \sqrt{9} = 3 \][/tex]
### Vertices
Vertices of the ellipse are located at [tex]\((\pm a, 0)\)[/tex]:
[tex]\[
(10, 0) \quad \text{and} \quad (-10, 0)
\][/tex]
So, the vertices are:
[tex]\[
[(10, 0), (-10, 0)]
\][/tex]
### Foci
The foci are located at [tex]\((\pm c, 0)\)[/tex], where [tex]\(c\)[/tex] is determined by:
[tex]\[
c = \sqrt{a^2 - b^2}
\][/tex]
First, calculate [tex]\(a^2 - b^2\)[/tex]:
[tex]\[
a^2 - b^2 = 100 - 9 = 91
\][/tex]
Then:
[tex]\[
c = \sqrt{91} \approx 9.539392014169456
\][/tex]
So, the foci are:
[tex]\[
(9.539392014169456, 0) \quad \text{and} \quad (-9.539392014169456, 0)
\][/tex]
Hence, the coordinates of the foci are:
[tex]\[
[(9.539392014169456, 0), (-9.539392014169456, 0)]
\][/tex]
### Endpoints of the Minor Axis
The endpoints of the minor axis are located at [tex]\((0, \pm b)\)[/tex]:
[tex]\[
(0, 3) \quad \text{and} \quad (0, -3)
\][/tex]
Thus, the endpoints of the minor axis are:
[tex]\[
[(0, 3), (0, -3)]
\][/tex]
### Final Results
- Vertices:
[tex]\[
[(10, 0), (-10, 0)]
\][/tex]
- Foci:
[tex]\[
[(9.539392014169456, 0), (-9.539392014169456, 0)]
\][/tex]
- Endpoints of the Minor Axis:
[tex]\[
[(0, 3), (0, -3)]
\][/tex]
