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 \square ) \\)
foci : \\( ( 0, \pm \square ) \\)

Explanation:

Step1: Identify the form of the ellipse equation

The standard form of an ellipse is \(\frac{x^{2}}{b^{2}}+\frac{y^{2}}{a^{2}} = 1\) (\(a>b>0\)) for a vertically - oriented ellipse. Given \(\frac{x^{2}}{144}+\frac{y^{2}}{169}=1\), we have \(b^{2}=144\), \(a^{2}=169\). Then \(a = 13\), \(b = 12\).

Step2: Find the endpoints of the major axis

For the ellipse \(\frac{x^{2}}{b^{2}}+\frac{y^{2}}{a^{2}}=1\), the endpoints of the major axis are \((0,\pm a)\). Since \(a = 13\), the endpoints of the major axis are \((0,\pm13)\).

Step3: Calculate the value of \(c\) (distance from the center to the foci)

Use the relationship \(c^{2}=a^{2}-b^{2}\). Substitute \(a^{2}=169\) and \(b^{2}=144\) into the formula: \(c^{2}=169 - 144=25\). Then \(c = 5\).

Step4: Find the foci

For the ellipse \(\frac{x^{2}}{b^{2}}+\frac{y^{2}}{a^{2}}=1\), the foci are located at \((0,\pm c)\). Since \(c = 5\), the foci are \((0,\pm5)\).

Answer:

The endpoints of the major axis: \(13\); The foci: \(5\)