Answer
Foci: $(-1,\pm\sqrt{21}+2)$
Vertices: $(-1,7)$ and $(-1,-3)$
Work Step by Step
Recall: The ellipse $\frac{(x-h)^2}{b^2}+\frac{(y-k)^2}{a^2}=1$ for $a\geq b$ has the foci at $(h,\pm c+k)$ and the vertices at $(h,\pm a+k)$
We have $25x^2+4y^2+50x-16y=59$.
Rewrite the equation above:
$25x^2+50x+25+4y^2-16y+16=59+25+16$
$25(x+1)^2+4(y-2)^2=100$
$\frac{(x+1)^2}{4}+\frac{(y-2)^2}{25}=1$
Then,
$h=-1,k=2,b=2$ and $a=5$
$c=\sqrt{a^2-b^2}=\sqrt{5^2-2^2}=\sqrt{25-4}=\sqrt{21}$
So, the foci are $(-1,\pm \sqrt{21}+2)$ and the vertices are $(-1,7)$ and $(-1,-3)$.
Sketch the graph:
