Question 449074
determine an equation for the hyberbola with center (0,-2), one vertex at (0,-4) and one foci at (0,2)
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Standard form for a hyperbola with horizontal transverse axis: (x-h)^2/a^2-(y-k)^2/b^2=1
Standard form for a hyperbola with vertical transverse axis: (y-k)^2/a^2-(x-h)^2/b^2=1
Note that the only difference between these two forms is that (x-h)^2 and (y-k)^2 are interchanged.
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Since given vertex is vertical, hyperbola is of the first form, that is, it has a vertical transverse axis.
a=distance from center to one of the vertices on the transverse axis=2
a^2=4
c=distance from center to one of the foci on the transverse axis=4
c^2=16
c^2=a^2+b^2
b^2=c^2-a^2=16-4=12
b=√12=3.46.. 
you now have enough information to write the equation for given hyperbola as follows:
ctr (0,-2)
a^2=4
b^2=12

(y+2)^2/4-(x-0)^2/12=1
(y+2)^2/4-x^2/12=1
See graph below as visual evidence of given hyperbola:
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y=(4(1+x^2/12))^.5-2
equation of asymptotes =±(4x/6.92)-2
{{{ graph( 300, 300, -10, 10, -10, 10, (4(1+x^2/12))^.5-2,-(4(1+x^2/12))^.5-2,(4x/6.92)-2,-(4x/6.92)-2) }}}