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