Answer
$\frac{(x-1)^2}{25}-\frac{(y-2)^2}{5}=1$
hyperbola: center $C(1,2)$, foci $F(1\pm\sqrt {30} ,2)$, vertices $V_1(6,2)$, $V_2(-4,2)$
asymptotes: $y=\pm\frac{\sqrt 5}{5}(x-1)+2$
See graph.
Work Step by Step
Step 1. Make squares of the variables: $(x^2-2x+1)-5(y^2-4y+4)=44+1-20$, $(x-1)^2-5(y-2)^2=25$. Divide 25 on both sides to get $\frac{(x-1)^2}{25}-\frac{(y-2)^2}{5}=1$
Step 2. We can identify the conic as a hyperbola centered at $C(1,2)$
Step 3. With $a=5, b=\sqrt 5$, we have $c=\sqrt {25+5}=\sqrt {30}$ and the foci $F(1\pm\sqrt {30} ,2)$
Step 4. The vertices can be found as $V(1\pm5 ,2)$ so that $V_1(6,2)$, $V_2(-4,2)$
Step 5. The original asymptotes are $y=\pm\frac{\sqrt 5}{5}x$ and the new asymptotes are $y=\pm\frac{\sqrt 5}{5}x$. With the shifts given by the new center, we have the new asymptotes as $y-2=\pm\frac{\sqrt 5}{5}(x-1)$ which gives $y=\pm\frac{\sqrt 5}{5}(x-1)+2$
Step 6. See graph.
