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Question 455268: x^2-y^2+6x+10y-17=0
A.) Write the equation in standard form.
B.) Find the Center, Verticies, Foci, and slope of asymptotes.
Answer by lwsshak3(11628)   (Show Source):
You can put this solution on YOUR website! x^2-y^2+6x+10y-17=0
A.) Write the equation in standard form.
B.) Find the Center, Verticies, Foci, and slope of asymptotes.
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Standard form of a hyperbola with horizontal transverse axis: (x-h)^2/a^2-(y-k)^2/b^2=1, with (h,k) being the (x,y) coordinates of the center.
Standard form of a hyperbola with vertical transverse axis: (y-k)^2/a^2-(x-h)^2/b^2=1, with (h,k) being the (x,y) coordinates of the center.
The difference between the two forms is the interchange of (x-h) and (y-k)
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x^2-y^2+6x+10y-17=0
completing the square
(x^2+6x+9)-(y^2-10y+25)=17+9+25=51
(x+3)^2/51-(y-5)^2/51=1
This is a hyperbola with horizontal transverse axis (first form listed above).
center:(-3,5)
a^2=51
a=√51=7.1..
vertices: (-3±√51,5)
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b^2=51
b=√51
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c^2=a^2+b^2=51+51=102
c=√(102)=10.1..
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foci: (-3±√(102),5)
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asymptotes
slope=±b/a=±√51/√51=±1
Equation of asymptotes
y=mx+b
y=±x+b
solving for b using coordinates of center which are on the asymptotes
5=±3+b
b=8 and 2
Equations:
y=-x+2
y=x+8
See graph below as a visual check on the answers
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y=((x+3)^2-51)^.5+5
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