College Algebra, Chapter 8, 8.2, Section 8.2, Problem 44

Determine the equation for the ellipse with eccentricity $\displaystyle \frac{\sqrt{3}}{2}$, foci on $y$-axis and length of major axis $4$.
The equation $\displaystyle \frac{x^2}{b^2} + \frac{y^2}{a^2} = 1$ has a horizontal major axis with forci $(0,\pm c)$ in which lies on $y$-axis. Where as
$c^2 = a^2 - b^2$ and the length of axis is $2a$. Also, the eccentricity is determined as $\displaystyle \frac{c}{a}$. So if $2a = 4$, then $a = 2$.
And if $\displaystyle \frac{c}{a} = \frac{\sqrt{3}}{2}$, then

$
\begin{equation}
\begin{aligned}
\frac{c}{2} &= \frac{\sqrt{3}}{2}\\
\\
a &= \sqrt{3}
\end{aligned}
\end{equation}
$

Thus,

$
\begin{equation}
\begin{aligned}
c^2 &= a^2 - b^2\\
\\
b^2 &= a^2 - c^2\\
\\
b^2 &= 12^2 - (\sqrt{3})^2\\
\\
b^2 &= 1\\
\\
b &= 1
\end{aligned}
\end{equation}
$

Therefore, the equation is
$\displaystyle \frac{x^2}{1^2} + \frac{y^2}{2^2} = 1 \text{ or } x^2 + \frac{y^2}{4} = 1$