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 equation of the ellipse is $\frac{x^{2}}{144}+\frac{y^{2}}{169}=1$, which is of the form $\frac{x^{2}}{b^{2}}+\frac{y^{2}}{a^{2}} = 1$ ($a>b$). Here $a^{2}=169$, so $a = 13$, and $b^{2}=144$, so $b = 12$.

Step2: Find the ends of the major - axis

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

Step3: Calculate the value of $c$

The relationship between $a$, $b$, and $c$ (semi - focal length) 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 an ellipse of the form $\frac{x^{2}}{b^{2}}+\frac{y^{2}}{a^{2}}=1$, the foci are located at $(0,\pm c)$. Since $c = 5$, the foci are $(0,\pm5)$.

Answer:

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