An ellipse intersects the $x$-axis at two points. One intersection point is the origin. If the foci of the ellipse are $(0,2)$ and $(3,2)$, then what is the other intersection point? Enter your answer as an ordered pair
Since one intersection point of the ellipse is $(0,0)$, we can see that the center of the ellipse is $(0,1)$. We know that $a=1$,
so we compute our answer as follows. Since the distance from $(0,0)$ to $(0,2)$ is 2 units, the distance from $(0,0)$ to $(0,1)$ is 1 unit. We also know that the distance between the two foci is $2c=3$, so $c=\frac{3}{2}$, which tells us $b^2=1^2-\left(\frac{3}{2}\right)^2=\frac{5}{4}$. Therefore, the equation of the ellipse is
\[\frac{x^2}{1^2}+\frac{y^2}{\frac{5}{4}}=1.\]To find the other $x$-intercept, we set $y$ to 0, and find that $x = \sqrt{1^2 - \left(\frac 35\right)^2} = \boxed{\left(\frac 45, 0\right)}$.[asy]
unitsize (1 cm);
draw((-2,0)--(5,0));
draw((0,-4/5)--(0,4));
draw(Circle((0,1),sqrt(5/4)));
dot("$(0,0)$", (0,0), SW);
dot("$(3,2)$", (3,2), NE);
dot("(0,2)", (0,2), N);
dot("$(0,1)$",(0,1),S);
dot("$(4/5,0)$",(4/5,0),S);
[/asy]