Answer
$\frac{(x-1)^2}{3}+\frac{(y-2)^2}{4}=1$
Work Step by Step
Step 1. Identify the given quantities: ellipse foci $F_1(1,1)$ and $F_2(1,3)$, one vertex on x-axis.
Step 2. The center is at the midpoint of the two foci, so we have $C(1, 2)$
Step 3. The distance between two foci is $2c=3-1=2$, thus $c=1$
Step 4. As one vertex is on the x-axis, we have $a=2$ (distance from the center to the x-axis)
Step 5. Use the relation $b^2=a^2-c^2=4-1=3$, we have $b=\sqrt 3$
Step 6. The major axis is vertical, and we can write the equation as $\frac{(x-1)^2}{3}+\frac{(y-2)^2}{4}=1$
