Answer
$\dfrac{x^2}{29}+\dfrac{y^2}{4}=1$
Work Step by Step
The standard form of an ellipse can be written as: $\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}=1....(1); $ and has foci $( \pm c, 0)$ where ; $c^2=a^2-b^2$
We are given that an ellipse with minor axis is $4$ . So, $2b=4 \implies b=2$ and $b^2 =4$.
Foci are: $(-5, 0)$ and $(5, 0)$; This implies that $c=5$
Therefore, $c^2=a^2-b^2 \implies 5^2 =a^2-2^2 \implies a^2= 29$
Now, substitute these values into the equation (1) to obtain:
$\dfrac{x^2}{29}+\dfrac{y^2}{4}=1$
