Answer
Hyperbola, center $C(0,0)$, foci $F(0,\pm12\sqrt 2)$, vertices $V(0,\pm12)$, asymptotes $y=\pm x$
See graph.
Work Step by Step
Step 1. Identify the type of the conics: rewrite the equation as $y^2-x^2=144$ which gives $\frac{y^2}{144}-\frac{x^2}{144}=1$, thus the equation is of the form of a hyperbola.
Step 2. The center can be found as $C(0,0)$
Step 3. With $a=12, b=12, c=12\sqrt 2$, the foci can be found as $F(0,\pm12\sqrt 2)$
Step 4. The vertices can be found as $V(0,\pm12)$
Step 5. The asymptotes can be found as $y=\pm x$
Step 6. See graph.
