College Algebra, Chapter 8, 8.2, Section 8.2, Problem 42

Determine the equation for the ellipse with eccentricity $\displaystyle \frac{1}{9}$ and foci $(0,\pm 2)$
The equation $\displaystyle \frac{x^2}{b^2} + \frac{y^2}{a^2} = 1$ is an ellipse with vertices major axis with foci $(0, \pm c)$
where $c^2 = a^2 - b^2$ and whose eccentricity is determined as $\displaystyle e = \frac{c}{a}$. So, $c = 2$ and if
$\displaystyle \frac{c}{a} = \frac{1}{9}$, then

$\begin{equation} \begin{aligned} \frac{2}{a} &= \frac{1}{9}\\ \\ a &= 18 \end{aligned} \end{equation}$

Thus,

$\begin{equation} \begin{aligned} c^2 &= a^2 - b^2\\ \\ b^2 &= a^2 - c^2\\ \\ b^2 &= 18^2 - 2^2\\ \\ b^2 &= 320\\ \\ b &= 8 \sqrt{5} \end{aligned} \end{equation}$

Therefore, the equation is
$\displaystyle \frac{x^2}{(8\sqrt{5})^2} + \frac{y^2}{(18)^2} = 1 \text{ or } \frac{x^2}{320} + \frac{y^2}{324} = 1$