Question 1049488
Vertices (0, ±4), asymptotes y = ± 1/2x
OPens Up an Down
Standard Form of an Equation of an Hyperbola opening up and down is:
  {{{(y-k)^2/b^2 - (x-h)^2/a^2 = 1}}} with C(h,k) and vertices 'b' units up and down from center,  2b the length of the transverse axis
Foci {{{sqrt(a^2+b^2)}}}units units up and down from center, along x = h
& Asymptotes Lines passing thru C(h,k), with slopes m =  ± b/a
Vertices (0, ±4) therefore C(0,0)
  {{{(y)^2/4^2 - (x)^2/a^2 = 1}}}
asymptotes y = ± 1/2x
m = ± b/a 
4/a = 1/2
a = 8
{{{(y)^2/4^2 - (x)^2/8^2 = 1}}}
{{{drawing(300,300,-20,20,-20,20, blue(line(0,10,0,-10)),
 grid(1),
circle(0,0,0.4),
circle(0, -4,0.4),
circle(0, 4,0.4),

graph( 300, 300,-20,20,-20,20,0, .5x, -.5x, 4sqrt((x)^2/64 + 1),-4sqrt((x)^2/64 +1 )  ))}}}