Question 721973: Find the center, vertices, and foci of the hyperbola and put the equation in standard form (x^2)-(9y^2)+2x-54y-107=0
Answer by lwsshak3(11628)   (Show Source):
You can put this solution on YOUR website! Find the center, vertices, and foci of the hyperbola and put the equation in standard form
(x^2)-(9y^2)+2x-54y-107=0
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complete the square:
(x^2)-(9y^2)+2x-54y-107=0
(x^2)+2x-(9y^2)-54y-107=0
(x^2+2x+1)-9(y^2+6y+9)=107+1+-81
(x+1)^2-9(y+3)^2=27
This is an equation of a hyperbola with horizontal transverse axis.
Its standard form:  , (h,k)=(x,y) coordinates of the center
For given hyperbola:
center: (-1,-3)
a^2=27
a=√27≈5.2
vertices: (-1±a,-3)=(-1±5.2,-3)=(-6.2,-3) and (4.2,-3)
..
c^2=a^2+b^2=27+3=30
c=√30≈5.5
foci:(-1±c,-3)=(-1±5.5,-3)=(-6.5,-3) and (4.5,-3)
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