Given the equation of the ellipse:
[tex]\[
\frac{x^2}{144} + \frac{y^2}{169} = 1
\][/tex]
We need to identify the lengths of the major axis and the locations of the foci.
First, recall the standard form of the ellipse equation:
[tex]\[
\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1
\][/tex]
By comparing this with the given equation:
[tex]\[
\frac{x^2}{144} + \frac{y^2}{169} = 1
\][/tex]
we can identify [tex]\( a^2 = 144 \)[/tex] and [tex]\( b^2 = 169 \)[/tex].
### Major Axis
For an ellipse, the length of the major axis is determined as follows:
- If [tex]\( a^2 < b^2 \)[/tex], the major axis is along the [tex]\( y \)[/tex]-axis.
Since [tex]\( 144 < 169 \)[/tex], the major axis is along the [tex]\( y \)[/tex]-axis.
The length of the semi-major axis [tex]\( b \)[/tex] is given by:
[tex]\[
b = \sqrt{169} = 13
\][/tex]
This means the ends of the major axis are at:
[tex]\[
(0, \pm b) = (0, \pm 13)
\][/tex]
### Foci
The foci of the ellipse are found using the relation:
[tex]\[
c = \sqrt{b^2 - a^2}
\][/tex]
where [tex]\( c \)[/tex] is the distance from the center to each focus.
Substitute [tex]\( b^2 = 169 \)[/tex] and [tex]\( a^2 = 144 \)[/tex]:
[tex]\[
c = \sqrt{169 - 144} = \sqrt{25} = 5
\][/tex]
Therefore, the foci are located at:
[tex]\[
(0, \pm c) = (0, \pm 5)
\][/tex]
### Summary
Therefore:
- The ends of the major axis are [tex]\((0, \pm 13)\)[/tex].
- The foci are located at [tex]\((0, \pm 5)\)[/tex].
To summarize:
[tex]\[
\text{Major axis } (0, \pm 13) \\
\text{Foci } (0, \pm 5)
\][/tex]
