Answer
Hyperbola, center $C(1,0)$, foci $F(1, \pm\sqrt 3)$, vertices $V(1, \pm\sqrt 2)$, asymptotes $y=\pm \sqrt 2(x-1)$
See graph.
Work Step by Step
Step 1. Identify the type of the conics: rewrite the equation as $2(x^2-2x+1)-(y^2)=2-4$ or $y^2-2(x-1)^2=2$ which gives $\frac{y^2}{2}-(x-1)^2=1$, thus the equation is of the form of a hyperbola.
Step 2. The center can be found as $C(1,0)$
Step 3. With $a=\sqrt 2, b=1, c=\sqrt {2+1}=\sqrt 3$, the foci can be found as $F(1, \pm\sqrt 3)$
Step 4. The vertices can be found as $V(1, \pm\sqrt 2)$
Step 5. The original asymptotes can be found as $y=\pm \sqrt 2(x)$ and the shifted asymptotes are $y=\pm \sqrt 2(x-1)$
Step 6. See graph.
