Question 509159: y^2-9x^2+2y+10=0 sketch the graoh,find the center,vertices,foci,eccentricity,asymptots.
Answer by lwsshak3(11628)   (Show Source):
You can put this solution on YOUR website! y^2-9x^2+2y+10=0 sketch the graph,find the center,vertices,foci,eccentricity,asymptotes.
**
complete the square
(y^2+2y+1)-9x^2=-10+1=-9
(y+1)^2-9x^2=-9
divide by -9
-(y+1)^2/9+x^2=1
x^2-(y+1)^2/9=1
This an equation of a hyperbola with horizontal transverse axis of the standard form:
(x-h)^2/a^2-(y-k)^2/b^2=1
For given equation:
Center: (0,-1)
a^2=1
a=1
vertices: (0±a,-1)=(0±1,-1)=(1,-1) and (-1,-1)
..
b^2=9
b=3
..
c^2=a^2+b^2=9+1=10
c=√10=3.16..
Foci: (0±c,-1)=(0±3.16,-1)=(3.16,-1) and (-3.16,-1)
..
Eccentricity: c/a=√10/1=√10=3.16
..
Asymptotes:
slope=±b/a=±3/1=±3
Equation: y=mx+b
For slope=3
y=3x+b
find b using coordinates of center (0,-1)
-1=3*0+b
b=-1
equation of asymptote: y=3x-1
..
For slope=-3
y=-3x+b
find b using coordinates of center (0,-1)
-1=3*0+b
b=-1
equation of asymptote: y=-3x-1
ans:
Center: (0,-1)
Vertices:(1,-1) and (-1,-1)
Foci: (3.16,-1) and (-3.16,-1)
Eccentricity: 3.16
Equation of asymptotes: y=3x-1 and y=-3x-1
..
see graph below:
..
y=±(9x^2-9)^.5-1
|
|
|
