Answer
See below
Work Step by Step
Given: $4x^2+9y^2+40x+72y+208=0$
Because $A=4,B=0$, and $C=9$, the discriminant is $B^2-4AC=-144\lt0$, so the conic is an ellipse. Complete the square to write the equation in standard form.
$4x^2+9y^2+40x+72y+208=0\\4(x^2+10x+25)-100+9(y^2+8y+16)-144+208=0\\4(x+5)^2+9(y+4)^2=36\\\frac{(x+5)^2}{9}+\frac{(y+4)^2}{4}=1$
From the equation, we can see that $(-5,-4)$ and $a=3,b=2$. The graph is shown below.
