College Algebra, Chapter 8, 8.2, Section 8.2, Problem 22

Determine the vertices, foci and eccentricity of the ellipse $20x^2+4y^2 = 5$. Determine the lengths of the major and minor
axes, and sketch the graph.
If we divide both sides by $5$, then we have
$\displaystyle 4x^2 + \frac{4y^2}{5} = 1$
Then, we can rewrite the equation as
$\displaystyle \frac{x^2}{\frac{1}{4}} + \frac{y^2}{\frac{5}{4}} = 1$

Notice that the equation 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 $\displaystyle a^2 = \frac{5}{4}$ and $\displaystyle b^2 = \frac{1}{4}$. So,
$\displaystyle c^2 = a^2 - b^2 = \frac{5}{4} - \frac{1}{4} = 1$. Thus, $\displaystyle a = \frac{\sqrt{5}}{2}, b = \frac{1}{2}$ and
$\displaystyle c = 1$. Then, the following is determined as

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