Answer
(a) center $C(-1, 3)$, vertices $V_1(-1,1)$ and $V_2(-1,5)$, foci $F(-1, 3\pm2\sqrt {10})$, asymptotes $y=\pm\frac{1}{3} (x+1)+3$
(b) See graph.
Work Step by Step
(a) Step 1. Write the given equation as $\frac{(y-3)^2}{4}-\frac{(x+1)^2}{36}=1$
Step 2. Identify the center as $C(-1, 3)$
Step 3. Identify $a=2, b=6, c=\sqrt {4+36}=2\sqrt {10}$
Step 4. The original vertices are $(0, \pm2)$, vertices after the shift $(-1, 3\pm2)$ or $V_1(-1,1)$ and $V_2(-1,5)$
Step 5. Original foci $(0, \pm2\sqrt {10})$, foci after shift $F(-1, 3\pm2\sqrt {10})$
Step 6. Original asymptotes $y=\pm\frac{1}{3} x$, new asymptotes $y=\pm\frac{1}{3} (x+1)+3$
(b) See graph.
