Question

find the ends of the major axis and foci of this ellipse. \\(\frac{x^2}{144} + \frac{y^2}{169} = 1\\) major axis : \\((0, \pm ?)\\) foci : \\((0, \pm \square)\\)

Explanation:

Step1: Identify ellipse form

The standard form of an ellipse centered at the origin with a vertical major axis is \(\frac{x^2}{b^2}+\frac{y^2}{a^2}=1\) where \(a > b\). Here, \(\frac{x^2}{144}+\frac{y^2}{169}=1\), so \(a^2 = 169\), \(b^2 = 144\).

Step2: Find \(a\) for major axis

Take square root of \(a^2\): \(a=\sqrt{169}=13\). So major axis ends are \((0, \pm 13)\).

Step3: Calculate \(c\) for foci

Use \(c^2=a^2 - b^2\). Substitute \(a^2 = 169\), \(b^2 = 144\): \(c^2=169 - 144 = 25\). Then \(c=\sqrt{25}=5\). So foci are \((0, \pm 5)\).

Answer:

Major axis: \((0, \pm 13)\)
Foci: \((0, \pm 5)\)