Question 874663
Transverse Axis is the segment connecting the vertices.  Yours is a length of 8 units, so a=4.  Distance from center to either vertex is 4 units, half of the transverse axis length.


Focal length:  {{{(7+abs(-5))/2=6=c}}}.  Again using focus information, the center, y value is the average of the focus y values:  {{{(7+(-5))/2=1}}}.
The center is (-3,1).


The negative sign is on the term for x, since your foci are vertically arranged.  If the hyperbola were standard position, vertices would be on the y axis.  In your case, center is NOT at the origin.


Now you have center, a, and c.  You can say,
{{{highlight_green((y-1)^2/4^2-(x+3)^2/b^2=1)}}}.
You still want the b value.


A substitution is often used in deriving the equation of a hyperbola.  
The effect is the relationship {{{a^2+b^2=c^2}}}
{{{b^2=c^2-a^2}}}.
You example has {{{b^2=6^2-4^2=36-16=20}}}.


Finally, your standard form hyperbola is {{{highlight((y-1)^2/4^2-(x+3)^2/20=1)}}}