Question 600340
find the foci, vertices, equations of the asymptotes, length of the transverse axis, and the length of the conjugate axis for the hyperbola 
16y^2-x^2=16
divide by 16
y^2-x^2/16=1
This is an equation of a hyperbola with vertical transverse axis (y-term listed ahead of x-term)
Its standard form: (y-h)^2/a^2-(y-k)^2/b^2=1, (h,k)=(x,y) coordinates of center
For given equation:
center: (0,0)
..
a^2=1
a=1
length of vertical transverse axis=2a=2
Vertices: (0,0±a)=(0,0±1)=(0,1) and (0,-1)
..
b^2=16
b=√16=4
length of conjugate axis=2b=8
..
Foci:
c^2=a^2+b^2=1+16=17
c=√17≈4.1
Foci: (0,0±c)=(0,0±4.1)=(0,4.1) and (0,-4.1)
..
Asymptotes:
Asymptotes are straight lines that go thru the center(0,0)
Equation: y=mx+b, m=slope, b=y-intercept
slopes:±a/b=±1/4
y=x/4+b
since asymptotes go thru the origin, b=0
Equations of asymptotes:
y=x/4
y=-x/4