Answer
$\frac{y^2}{16}-\frac{(x-3)^2}{9}=1$
center $C(3,0)$, foci $F(3,\pm5)$, vertices $V(3,\pm4)$
asymptotes: $y=\pm\frac{4}{3}(x-3)$
See graph.
Work Step by Step
Step 1. Form square with variables: $16(x^2-6x+9)-9y^2=144-288$ or $16(x-3)^2-9y^2=-144$. Divide both sides by $-144$, we get $\frac{y^2}{16}-\frac{(x-3)^2}{9}=1$
Step 2. The above equation shows a hyperbola with a center at $C(3,0)$
Step 3. With $a=4, b=3$, we have $c=\sqrt {16+9}=5$ and the foci are $F(3,\pm5)$
Step 4. The vertices are at $V(3,\pm4)$
Step 5. The original asymptotes are $y=\pm\frac{4}{3}x$, with the center shifted to $(3,0)$, the new asymptotes are $y=\pm\frac{4}{3}(x-3)$
Step 6. See graph.
