What are the vertices, foci and asymptotes of the hyperbola with equation 16x^2-4y^2=64

The hyperbola: x^2/a^2 - y^2/b^2 = 1 
foci: (+/-c,0) = (+/-sqrt(5),0) where c^2 = 1^2+2^2 
and vertices (+/-a,0) and 
asimptotes: y = +/-(b/a)x so, 

16x^2-4y^2=64 
x^2/1^2 - y^2/2^2 = 4 
here, a = 1; b = 2 and c = sqrt[(a^2+b^2)] = sqrt[1 + 4] = sqrt(5) 

So, 16x^2-4y^2=64 has 
foci: foci: (+/-c,0) = (+/-sqrt(5),0) and 
vertices: (+/-a,0) = (+/-1,0) and 
asimptotes: y = +/-(b/a)x = +/-(2/1)x = +/-2x

Absolute value graph shifted point 5 up