Answer
$\frac{(x-3)^2}{9}+\frac{y^2}{25}=1$
Work Step by Step
Step 1. As the foci are on the y-axis, assume the ellipse has a general form $\frac{(x-h)^2}{b^2}+\frac{(y-k)^2}{a^2}=1$
Step 2. Based on the foci coordinates, we have $2c=4+4$ and $c=4$
Step 3. Based on the foci positions, we can identify the center as $(3,0)$, thus $h=3,k=0$
Step 4. The x-intercepts gives points on the curve $(0,0)$ and $(6,0)$
Step 5. Plug-in one of the above point to the equation, we have $\frac{(0-3)^2}{b^2}+\frac{(0)^2}{a^2}=1$ which gives $b^2=9$
Step 6. Use the relationship $a^2=b^2+c^2$, we have $a^2=9+16=25$
Step 7. Conclusion: the ellipse equation is $\frac{(x-3)^2}{9}+\frac{(y)^2}{25}=1$
