Answer
asymptotes: $y=\pm\frac{2}{3}(x-2)+3$
foci: $(2\pm\sqrt {13},3)$
Work Step by Step
Step 1. Rewriting the equation as
$4(x^2-4x+4)-9(y^2-6y+9)=101+16-81=36$
or
$\frac{(x-2)^2}{9}-\frac{(y-3)^2}{4}=1$
we have $a=3,b=2, c=\sqrt {a^2+b^2}=\sqrt {13}$ centered at $(2,3)$ with a horizontal transverse axis.
Step 2. We can find the vertices as $(-1,3),(5,3)$ and asymptotes as $y-3=\pm\frac{b}{a}(x-2)$ or $y=\pm\frac{2}{3}(x-2)+3$
Step 3. We can graph the equation as shown in the figure with foci at $(2\pm\sqrt {13},3)$
