Answer
$\frac{x^2}{9}+\frac{(y-4)^2}{25}=1$
Work Step by Step
Step 1. Identify the given quantities: ellipse center $C(0,4)$, foci $F_1(0,0)$ and $F_2(0,8)$, major axis of length $10$.
Step 2. With the length of the major axis, we have $2a=10$ and $a=5$
Step 3. As the foci are along the vertical direction, we have a general equation as $\frac{(x-h)^2}{b^2}+\frac{(y-k)^2}{a^2}=1$
Step 4. With the center at $(0,4)$, we have $h=0, k=4$
Step 5. The distance between the focal points $2c=8-0=8$ and $c=4$
Step 6. Use the relation $b^2=a^2-c^2=25-16=9$, thus $b=3$
Step 7. Write the equation for the ellipse as $\frac{(x-0)^2}{3^2}+\frac{(y-4)^2}{5^2}=1$, and we get the answer as $\frac{x^2}{9}+\frac{(y-4)^2}{25}=1$
