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Question 511168: A hyperbola has vertices at (-3,3) and (5,3) and a focus at (7,3).
What is the equation for this hyperbola?
write an equation for one of the assymptotes
Answer by lwsshak3(11628)   (Show Source):
You can put this solution on YOUR website! A hyperbola has vertices at (-3,3) and (5,3) and a focus at (7,3).
What is the equation for this hyperbola?
write an equation for one of the asymptotes
**
As gleaned from given information(draw a rough sketch with given points), this hyperbola has a horizontal transverse axis with an equation of the standard form:
(x-h)^2/a^2-(y-k)^2/b^2=1
For given hyperbola:
center:(1,3)
Vertices: (-3,3) and (5,3) (given)
..
length of transverse axis=5+3=8=2a (distance between end points of vertex)
a=4
a^2=16
..
c=distance from center to focus on horizontal transverse axis=7-1=6
c^2=36
..
c^2=a^2+b^2
b^2=c^2-a^2=36-16=20
b=√20≈4.47
..
Asymptotes:
slope=±b/a=±√20/4=1.118
Equations of asymptotes are straight lines of the standard form: y=mx+b, m=slope,
b=y-intercept, which pass thru the center of the hyperbola.
For asymptote with slope>0
y=1.118x=b
solving for b, using coordinates of center
3=1.118*1+b
b=3-1.118=1.882
Equation of one asymptote: y=1.118x+1.882
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Equation of given hyperbola:
(x-1)^2/16-(y-3)^2/20=1
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