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 )\\)

Explanation:

Step1: Identify the form of the ellipse

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

Step2: Find the ends of the major axis

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

Step3: Calculate the value of c

The relationship for an ellipse is $c^{2}=a^{2}-b^{2}$. Substituting $a^{2}=169$ and $b^{2}=144$, we get $c^{2}=169 - 144=25$, so $c = 5$.

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:

Major axis: $(0,\pm13)$
Foci: $(0,\pm5)$