Answer
$\frac{(x+2)^2}{8}-\frac{(y-4)^2}{9}=1$, hyperbola, center $C(-2, 4)$. foci $F(-2\pm\sqrt {17}, 4)$, vertices $V(-2\pm2\sqrt {2}, 4)$, asymptotes $y=\pm\frac{3\sqrt 2}{4}(x+2)+4$
See graph.
Work Step by Step
Step 1. Rewrite the equation as $9(x^2+4x+4)-8(y^2-8y+16)=164+36-8\times16$ or
$9(x+2)^2-8(y-4)^2=72$
Step 2. Divide both sides by 72 to get $\frac{(x+2)^2}{8}-\frac{(y-4)^2}{9}=1$ which represents a hyperbola with $a=2\sqrt 2, b=3$.
Step 3. The center can be found at $C(-2, 4)$.
Step 4. To find the foci, use the relation $c^2=a^2+b^2$ to get $c=\sqrt {8+9}=\sqrt {17}$
Step 5. The original foci are $(\pm\sqrt {17}, 0)$, and the shifted foci are $F(-2\pm\sqrt {17}, 4)$
Step 6. The original vertices are $(\pm2\sqrt 2, 0)$, and the shifted vertices are $V(-2\pm2\sqrt {2}, 4)$
Step 7. The original asymptotes are $y=\pm\frac{b}{a}x=\pm\frac{3\sqrt 2}{4}x$, and the shifted asymptotes are $y=\pm\frac{3\sqrt 2}{4}(x+2)+4$
Step 8. See graph.
