Question 1067781
<pre><b>
{{{drawing(800,480,-15,5,-9,3,graph(800,480,-15,5,-9,3, (-1/3)sqrt(x^2+10x+61)-3),
graph(800,480,-15,5,-9,3, (1/3)sqrt(x^2+10x+61)-3),
circle(-5,-3,.1),locate(-5,-3,"(-5,-3)"), 
circle(-5,-5,.1),locate(-5,-5,"(-5,-5)"), 
circle(-5,-1,.1),locate(-5,-1,"(-5,-1)"), 
circle(-11,-3,.1),locate(-11,-3,"(11,-3)"), 
circle(1,-3,.1),locate(1,-3,"(1,-3)"),
green(line(-11,-5,1,-5),line(1,-5,1,-1),line(1,-1,-11,-1),line(-11,-1,-11,-5),
line(13,3,-17,-7),line(-17,1,7,-7)),
blue(triangle(-11,-3,1,-3,2,-3)), red(triangle(-5,-1,-5,-5,-5,-3))


 )}}}

The equation of a hyperbola that opens upward and downward has
general equation

{{{(y-k)^2/a^2-(x-h)^2/b^2}}}{{{""=""}}}{{{1}}}

where (h,k) is the center, a = half the transverse axis,
b = half the conjugate axis.

(h,k) = (-5,-3)

The red vertical line is the transverse axis. It is 4 units long.
The semi-transverse axis is half the transverse axis, so a=2

The blue horizontal line is the conjugate axis. It is 12 units long.
The semi-transverse axis is half the conjugate axis, so b=6

So the equation of the hyperbola, which is what you want, is

{{{(y-(-3)^"")^2/2^2-(x-(-5)^"")^2/6^2}}}{{{""=""}}}{{{1}}}

{{{(y+3)^2/4^""-(x+5)^2/36^""}}}{{{""=""}}}{{{1}}}

The green rectangle is the "defining rectangle".  The slanted green
lines are the extended diagonals of the defining rectangle.  They are 
the asymptotes of the hyperbola.

Edwin</pre></b>