Answer
asymptotes: $y=\pm\frac{2}{5}(x-4)$
foci: $(4,\pm\sqrt {29})$
Work Step by Step
Step 1. Rewriting the equation as
$4(x^2-8x+16)-25(y^2)=64-164=-100$
or
$\frac{(y)^2}{4}-\frac{(x-4)^2}{25}=1$
we have $a=2,b=5, c=\sqrt {a^2+b^2}=\sqrt {29}$ centered at $(4,0)$ with a vertical transverse axis.
Step 2. We can find the vertices as $(4,-2),(4,2)$ and asymptotes as $y=\pm\frac{a}{b}(x-4)$ or $y=\pm\frac{2}{5}(x-4)$
Step 3. We can graph the equation as shown in the figure with foci at $(4,\pm\sqrt {29})$
