The ellipse $9x^2+25y^2=225$ has foci $F'$ and $F$ on the $x$-axis, with $F$ on the right. The line through $F$ perpendicular to the $x$-axis meets the ellipse at $A$ (above the axis) and $B$. What is the distance $AF'$?

1. For $9x^2+25y^2=225$, i.e. $\frac{x^2}{25}+\frac{y^2}9=1$: $a=5$, $b=3$, $c=\sqrt{25-9}=4$, so the foci are $F'=(-4,0)$ and $F=(4,0)$.
2. At $x=4$: $25y^2=225-144=81$, so $y=\frac95$ and $A=\left(4,\frac95\right)$; thus $AF=\frac95$.
3. For any point of the ellipse, the distances to the two foci add up to $2a=10$. So $AF'=10-\frac95=\boxed{\dfrac{41}5}$.