Answer
$\dfrac{(x-5)^2}{25}+\dfrac{(y-2)^2}{16}=1$
Work Step by Step
The standard form of an ellipse can be written as: $\dfrac{(x-h)^2}{a^2}+\dfrac{(y-k)^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 $8$ . So, $2b=8 \implies b=4$ and $b^2 =16$.
Foci are: $(-3, 0)$ and $(3, 0)$; This implies that $c=3$
Therefore, $c^2=a^2-b^2 \implies 3^2 =a^2-4^2 \implies a^2= 25$
Now, substitute these values into the equation (1) to obtain:
$\dfrac{(x-5)^2}{25}+\dfrac{(y-2)^2}{16}=1$
