Um, actually there's one little error;
It's $\displaystyle \frac{x^2}{9}+\frac{y^2}{25}=1$ not $\displaystyle \frac{x^2}{25}+\frac{y^2}{9}=1$ because it's an ellipse streched along the y-axis. Look at page 420, they show two guidelines (in grey); the one on the left shows the guideline for an ellipse having center (0, 0),
foci (0, -c) and (0, c). You followed the guideline for an ellipse having center (0, 0),
foci (c, 0) and (c, 0).
I'll show ya;
"The ellipse having center (0, 0), foci (0, -c) and (0, c), and sum of focal radii 2a has the equation; ..............$\displaystyle \frac{x^2}{b^2}+\frac{y^2}{a^2}=1$ where $\displaystyle b^2 = a^2 - c^2$ "
So, to go back to the orginal question;
What is the equation of the ellipse with foci at (0,-4) and (0,4) ant the sum of its focal radii being 10?
So 10 = 2a, so a = 5 and c = 4. Now plug those values in $\displaystyle b^2 = a^2 - c^2$ ;
$\displaystyle 5^2 - 4^2 = b^2$
$\displaystyle 25 - 16 = 9$ so b = 3
Since the foci are on the y-axis, the major axis is vertical, so use the appropriate formula;
$\displaystyle \frac{x^2}{b^2}+\frac{y^2}{a^2}=1$ plug in the values;
$\displaystyle \frac{x^2}{3^2}+\frac{y^2}{5^2}=1$
$\displaystyle \frac{x^2}{9}+\frac{y^2}{25}=1$
Sorry if this is a little lengthy, I just wanted to make sure for the both of us. I actually erased my answers and wrote yours when I saw that ours were different
but then I double checked
Hoorah~