Answer
$\dfrac{x^2}{29}+\dfrac{y^2}{4}=1$
Work Step by Step
The ellipse has a minor axis with length of $4$ units.
This means that $2b=4$ which implies that $b=2$.
The foci are ar $(-5, 0)$ and $(5, 0)$, which implies that $c=5$.
The standard form of an ellipse whose foci are at $(-c, 0)$ and $(c, 0)$ is
$$\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}=1$$
and has its center at $(0, 0)$ (which is the origin)
With $b=2$ and $c=5$, use the formula $c^2=a^2-b^2$ to have:
$$c^2=a^2-b^2\\
5^2 =a^2-2^2\\
25=a^2-4\\
25+4=a^2\\
29=a^2$$
Therefore, standard form of the equation's ellipse is:
$$\dfrac{x^2}{29}+\dfrac{y^2}{4}=1$$
