Answer
(a) center $(0,-3)$, vertices $(\pm3,-3)$, foci $(\pm\sqrt {34},-3)$.
asymptotes: $y=\pm\frac{5}{3}x-3$.
(b) See graph.
Work Step by Step
(a) Step 1. Make square of the variables: $25x^2-9(y^2+6y)=306$ or $25x^2-9(y+3)^2=306-9\times3^2=15^2$. Divide $15^2$ on both sides to get $\frac{x^2}{3^2}-\frac{(y+3)^2}{5^2}=1$
Step 2. The un-shifted equation of the above hyperbola is $\frac{x^2}{3^2}-\frac{y^2}{5^2}=1$ and we can identify $a=3, b=5, c=\sqrt {3^2+5^2}=\sqrt {34}$, center $(0,0)$, vertices $(\pm3,0)$, foci $(\pm\sqrt {34},0)$, asymptotes $y=\pm\frac{5}{3}x$
Step 3. The new hyperbola will shift 3 units down from the one in step-2, and we get new center $(0,-3)$, vertices $(\pm3,-3)$, foci $(\pm\sqrt {34},-3)$.
Step 4. For the new asymptotes, keep the slopes and use point $(0,-3)$ to find the new y-intercepts. For $y=\frac{5}{3}x+b$, plug-in $(0,-3)$ we have $b=-3$ which gives $y=\frac{5}{3}x-3$. For $y=-\frac{5}{3}x+c$, plug-in $(0,-3)$ we have $c=-3$ which gives $y=-\frac{5}{3}x-3$.
(b) See graph.
