Solution: The standard form of the ellipse centered at the origin is $\fracx^2a^2 + \fracy^2b^2 = 1$, where $a > b$. Given the major axis is $10$, we have $2a = 10 \Rightarrow a = 5$. The minor axis is $6$, so $2b = 6 \Rightarrow b = 3$. The distance from the center to a focus is given by $c = \sqrta^2 - b^2$. Substituting $a = 5$ and $b = 3$, we get $c = \sqrt25 - 9 = \sqrt16 = 4$. - Nelissen Grade advocaten