Question 864674
For the hyperbola represented by [(y+2)^2]/4 - [x^2]/12=1 find the foci and vertices
{{{(y+2)^2/4 - x^2/12=1}}}
Given hyperbola has a vertical transverse axis.
Its standard form of equation: {{{(y-k)^2/a^2-(x-h)^2/b^2=1}}}, (h,k)=coordinates of center
for given hyperbola:
center: (0,-2)
a^2=4
a=2
vertices: (0,-2±a)=(0,-2±2)=(0,-4) and (0,0)
b^2=12
c^2=a^2+b^2=4+12=16
c=√16=4
foci:  (0,-2±c)=(0,-2±4)=(0,-6) and (0,2)