College Algebra, Chapter 8, 8.2, Section 8.2, Problem 18

Determine the vertices, foci and eccentricity of the ellipse $9x^2 + 4y^2 = 1$. Determine the lengths of the major and minor
axes, and sketch the graph.
If we divide both sides by $16$, then we have
$\displaystyle \frac{x^2}{\frac{1}{9}} + \frac{y^2}{\frac{1}{4}} = 1$
We'll see that the function has the form $\displaystyle \frac{x^2}{b^2} + \frac{y^2}{a^2} = 1$. Since the denominator of $y^2$ is larger, then the ellipse
has a vertical major axis. This gives $\displaystyle a^2 = \frac{1}{4}$ and $\displaystyle b^2 = \frac{1}{9}$. So,
$\displaystyle c^2 = a^2 - b^2 = \frac{1}{4} - \frac{1}{9} = \frac{5}{36}$. Thus, $\displaystyle a = \frac{1}{2}, b = \frac{1}{3}$ and
$\displaystyle c = \frac{\sqrt{5}}{6}$. Then, the following is determined as

$\begin{equation} \begin{aligned} \text{Vertices}& &(0, \pm a) &\rightarrow \left(0, \pm \frac{1}{2}\right)\\ \\ \text{Foci}& &(0, \pm c) &\rightarrow \left(0, \pm \frac{\sqrt{5}}{6}\right)\\ \\ \text{Eccentricity (e)}& &\frac{c}{a} &\rightarrow \frac{\sqrt{5}}{3}\\ \\ \text{Length of major axis}& &2a &\rightarrow 1\\ \\ \text{Length of minor axis}& &2b &\rightarrow \frac{2}{3} \end{aligned} \end{equation}$