Question 614750: Find the vertices, foci, asymtopes for y.
(y-2)^2/1 - (x-1)^2/4 =1
Answer by lwsshak3(11628)   (Show Source):
You can put this solution on YOUR website! Find the vertices, foci, asymptotes for y.
(y-2)^2/1 - (x-1)^2/4 =1
This is an equation of a hyperbola with vertical transverse axis.
Its standard form: (x-h)^2/a^2-(y-k)^2/b^2=1, (h,k)=(x,y) coordinates of the center.
For given equation: (y-2)^2/1 - (x-1)^2/4 =1
center: (1,2)
a^2=1
a=1
vertices: (1,2±a)=(1,2±1)=(1,1) and (1,3)
..
b^2=4
b=2
..
c^2=a^2+b^2=1+4=5
c=√5≈2.3
foci: (1,2±c)=(1,2±2.3)=(1,-.3) and (1,4.3)
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Both asymptotes are straight lines which go thru and intersect at the center.
Their standard form of equation: y=mx+b, m=slope, b=y-intercept
Slopes of asymptotes with vertical transverse axis=±a/b=±1/2
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Equation of asymptote with negative slope:
y=-x/2+b
solving for b using (x,y) coordinates of center(1,2)
2=-1/2+b
b=5/2
equation:y=-x/2+5/2
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Equation of asymptote with positive slope:
y=x/2+b
solving for b using (x,y) coordinates of center(1,2)
2=1/2+b
b=3/2
equation:y=x/2+3/2
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