Answer
$e=\dfrac{1}{2}$
Work Step by Step
The given ellipse has $a^2=4$ and $b^2=3$.
Solve for $a$ and $b$ to obtain:
\begin{align*}
a^2&=4\\
a&=\sqrt4\\
a&=2 \quad \quad\text{(since $a$ is non-negative, we only take the principal root)}\\
\\
b^2&=3\\
b&=\sqrt3 \quad \quad \text{(since $b$ is non-negative, we only take the principal root)}
\end{align*}
Solve for $c$ using the formula $c^2=a^2-b^2$ to obtain:
\begin{align*}
c^2 &=4-3\\
c^2&=1\\
c&=\sqrt1\quad \quad \text{(since $c$ is non-negative, we only take the principal root)}\\
c&=1
\end{align*}
The eccentricity $e$ is given by the formula $e=\dfrac{c}{a}$.
Thus, $e=\dfrac{1}{2}$.
