SOLUTION: find the foci, vertices, equations of the asyptotes, length of the transverse axis, and the length of the conjugate axis for the hyperbola 16y^2-x^2=16.

Question 600340: find the foci, vertices, equations of the asyptotes, length of the transverse axis, and the length of the conjugate axis for the hyperbola 16y^2-x^2=16.
Answer by lwsshak3(11628) (Show Source): You can put this solution on YOUR website!
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
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