Answer
See below
Work Step by Step
Given: $\frac{x^2}{4}+y^2=25\\\frac{x^2}{100}+\frac{y^2}{25}=1$
The equation is in standard form.
We can see $a=10, b=5$
The denominator of the $x^2-term$ is greater than that of the $y^2-term$, so the major axis is horizontal.
The vertices of the ellipse are at $(\pm a,0)=(\pm 10,0)$. The co-vertices are at $(0, \pm b) = (0, \pm 5)$. Find the foci.
$c^2=a^2-b^2=10^2-5^2=75$
So $c=5\sqrt 3$
The foci are at $(\pm 5\sqrt 3 , 0)$.
