Answer
hyperbola, center $(1,2)$, vertices $(1,0),(1,4)$, foci $(1,2-\sqrt {5}),(1,2+\sqrt {5})$, and asymptotes $y-2=\pm2(x-1)$.
Work Step by Step
1. Given $y^2-4y-4x^2+8x=4$ or $\frac{(y-2)^2}{4}-\frac{(x-1)^2}{1}=1$, we can identify it as a hyperbola with a vertical transverse axis, $a=2, b=1$ and $c=\sqrt {a^2+b^2}=\sqrt {5}$.
2. We can find its center $(1,2)$, vertices $(1,0),(1,4)$, foci $(1,2-\sqrt {5}),(1,2+\sqrt {5})$, and asymptotes $y-2=\pm\frac{a}{b}(x-1)=\pm2(x-1)$.
