Answer
See below
Work Step by Step
Given: $4x^2+y^2=36\\\frac{x^2}{9}+\frac{y^2}{36}=1$
The equation is in standard form.
We can see $a=6, b=3$
The denominator of the $x^2-term$ is smaller than that of the $y^2-term$, so the major axis is vertical.
The vertices of the ellipse are at $(0,\pm a)=(0,\pm 6)$. The co-vertices are at $(\pm b,0) = (\pm 3,0)$. Find the foci.
$c^2=a^2-b^2=6^2-3^2=27$
so $c=3\sqrt 3$
The foci are at $(0,\pm 3\sqrt 3)$.
