Answer
See below
Work Step by Step
Given: $\frac{x^2}{144}+\frac{y^2}{81}=1$
The equation is in standard form.
We can see $a=12, b=9$
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 $(\pm a,0)=(\pm 12,0)$. The co-vertices are at $(0,\pm b) = (0,\pm 9)$. Find the foci.
$c^2=a^2-b^2=12^2-9^2=80$
so $c=4\sqrt 5$
The foci are at $(\pm 4\sqrt 5,0)$.
