Question

find the center and foci of the ellipse.
\\(\frac{(x + 2)^{2}}{289}+\frac{(y - 4)^{2}}{225}=1\\)
center = (?, )
foci = (, ) and (, )

Explanation:

Step1: Recall ellipse standard form

The standard - form of an ellipse is $\frac{(x - h)^2}{a^2}+\frac{(y - k)^2}{b^2}=1$ for a horizontal ellipse (where $a>b$), and its center is $(h,k)$. Given $\frac{(x + 2)^2}{289}+\frac{(y - 4)^2}{225}=1$, we can rewrite it as $\frac{(x-(-2))^2}{17^2}+\frac{(y - 4)^2}{15^2}=1$.

Step2: Find the center

Comparing with the standard form, we have $h=-2$ and $k = 4$. So the center of the ellipse is $(-2,4)$.

Step3: Calculate $c$

For an ellipse, the relationship between $a$, $b$, and $c$ (distance from the center to the foci) is $c^2=a^2 - b^2$. Here, $a = 17$ and $b = 15$, so $c^2=17^2-15^2=(17 + 15)(17 - 15)=32\times2 = 64$, then $c = 8$.

Step4: Find the foci

Since the ellipse is horizontal (because $a^2=289$ is under the $x$ - term), the foci are given by $(h\pm c,k)$. Substituting $h=-2$, $k = 4$, and $c = 8$, we get the foci as $(-2+8,4)$ and $(-2 - 8,4)$, which are $(6,4)$ and $(-10,4)$.

Answer:

Center = $(-2,4)$
Foci = $(6,4)$ and $(-10,4)$