SOLUTION: One of the asymptotes of a hyperbola with a vertical transverse axis is y-4 = 2/3(x-1) The distance between the two vertices is 8 units. What is the equation of the hyperbola and w
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-> SOLUTION: One of the asymptotes of a hyperbola with a vertical transverse axis is y-4 = 2/3(x-1) The distance between the two vertices is 8 units. What is the equation of the hyperbola and w
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Question 458612: One of the asymptotes of a hyperbola with a vertical transverse axis is y-4 = 2/3(x-1) The distance between the two vertices is 8 units. What is the equation of the hyperbola and what are the coordinates of the two foci? Answer by lwsshak3(11628) (Show Source):
You can put this solution on YOUR website! One of the asymptotes of a hyperbola with a vertical transverse axis is y-4 = 2/3(x-1) The distance between the two vertices is 8 units. What is the equation of the hyperbola and what are the coordinates of the two foci
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Standard form of hyperbola with vertical transverse axis: (y-h)^2/a^2-(x-h)^2/b^2=1, with (h,k) being the (x,y) coordinates of the center.
length of vertical transverse axis=8=2a
a=4
a^2=16
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asymptotes:
y-4 = 2/3(x-1)
y=(2/3)x-2/3+4
y=(2/3)x-2/3+12/3
y=(2/3)x+10/3 (given asymptote)
and
y=-(2/3)x+10/3 (other asymptote)
y-intercept=10/3=y-coordinate of center
x-coordinate of center=0
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slope,m=2/3=a/b
b=(3/2)a=(3/2)4=6
b^2=36
Equation: (y-10/3)^2/16-x^2/36=1
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Foci:
c^2=a^2+b^2=16+36=52
c=√52=7.21..
coordinates of Foci:(0,10/3±√52)
see graph below as a visual check on the answers
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y=(16+16x^2/36)^.5+10/3
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