Answer
asymptotes $y=\pm4(x+2)-1$
foci: $(-2,-1\pm\frac{\sqrt {17}}{2})$
Work Step by Step
Step 1. Rewriting the equation as
$16(x^2+4x+4)-(y^2+2y+1)=64-1-67=-4$
or
$\frac{(y+1)^2}{4}-\frac{(x+2)^2}{1/4}=1$
we have $a=2,b=\frac{1}{2}, c=\sqrt {a^2+b^2}=\frac{\sqrt {17}}{2}$ centered at $(-2,-1)$ with a vertical transverse axis.
Step 2. We can find the vertices as $(-2,-3),(-2,1)$ and asymptotes as $y+1=\pm\frac{a}{b}(x+2)$ or $y=\pm4(x+2)-1$
Step 3. We can graph the equation as shown in the figure with foci at $(-2,-1\pm\frac{\sqrt {17}}{2})$
