Find the ends of the major axis and foci.

[tex]\[ \frac{x^2}{144} + \frac{y^2}{169} = 1 \][/tex]

Major axis: [tex]\((0, \pm 13)\)[/tex]

Foci: [tex]\((0, \pm \sqrt{169 - 144}) = (0, \pm 5)\)[/tex]



Answer :

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]