Question 309219
The center of the hyperbola is the midpoint of the line segment from vertex to vertex. So the midpoint of (0,6) and (0,-6) is ((0+0)/2, (6+(-6)/2) ---> (0, 0). So the center is (0,0). The center is in the form (h,k), so h=0 and k=0.


In this case, 'a' is the length of the semi-minor axis and 'b' is the length of the semi-major axis. So the lengths of the conjugate and traverse axes are 2a and 2b units respectively.


Since the traverse axis is 2b units long, this means that 2b=12 (since the distance between the vertices is 12 units) which means that b=6. Also, because the conjugate axis is 14 units long, and 2a is the length of the conjugate axis, this means that 2a=14 and a=7



Finally, recall that the general equation for a hyperbola which opens up vertically is {{{((y-k)^2)/(b^2)-((x-h)^2)/(a^2)=1}}}. From here, all you need to do is plug in the right values and simplify.