Question 476629: find the center, the vertices, foci, and the asymptotes of the hyperbola. then draw the graph.
x^2-y^2+4x+6y-6=0
Answer by lwsshak3(11628)   (Show Source):
You can put this solution on YOUR website! find the center, the vertices, foci, and the asymptotes of the hyperbola. then draw the graph.
x^2-y^2+4x+6y-6=0
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x^2-y^2+4x+6y-6=0
completing the squares:
(x^2+4x+4)-(y^2-6y+9)-6-4+9=0
(x+2)^2-(y-3)^2=1
This is an equation of a hyperbola with horizontal transverse axis of the standard form:
(x-h)^2/a^2-(y-k)^2=1
..
For given equation:
center:(-2,3)
a^2=1
a=1
length of transverse axis=2a=2
vertices: (-2±a,3) =(-2±1,3) or (-3,3) and (-1,3)
b^2=1
b=1
c^2=a^2+b^2=1+1=2
c=√2
Foci: (-2±c,3)=(-2±√2,3) or (-2-√2,3) and (-2+√2,3)
..
Asymptotes:
slope,m=±b/a=±1/1=±1
Equations:
y=±x+b
finding b using (x,y) coordinates of center (-2,3)
3=-2=b
b=5
equation: y=x+5
..
y=-x+b
3=-(-2)+b
b=1
equation: y=-x+1
..
center:(-2,3)
vertices: (-3,3) and (-1,3)
Foci: (-2-√2,3) and (-2+√2,3)
Asymptotes:y=x+5 and y=-x+1
see graph below as a visual check on answers:
..
y=±((x+2)^2-1)^.5+3
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