Answer
The foci are $(\pm \sqrt{21},1)$ and the vertices are $(\pm 5,1)$
Work Step by Step
Recall: The ellipse $\frac{(x-h)^2}{a^2}+\frac{(y-k)^2}{b^2}=1$ for $a\geq b$ has the foci at $(\pm c+h,k)$ and the vertices at $(\pm a+h,k)$.
We have
$4x^2+25y^2-50y=75$
$4x^2+25y^2-50y+25=75+25$
$4x^2+25(y^2-2y+1)=100$
$4x^2+25(y-1)^2=100$
$\frac{4x^2+25(y-1)^2}{100}=1$
$\frac{x^2}{25}+\frac{(y-1)^2}{4}=1$
$\frac{(x-0)^2}{5^2}+\frac{(y-1)^2}{2^2}=1$
Then,
$h=0$, $k=1$, $a=5$, and $b=2$
$c=\sqrt{a^2-b^2}=\sqrt{5^2-2^2}=\sqrt{21}$
So, the foci are $(\pm \sqrt{21},1)$ and the vertices are $(\pm 5,1)$.
