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Question 389754: -x^2+y^2+4x-13=0
classify each conic section and sketch it on a graph
Answer by lwsshak3(11628)   (Show Source):
You can put this solution on YOUR website! -x^2+y^2+4x-13=0
complete the square.
-(x^2+4x+4) +y^2=13-4
-(x+2)^2+y^2 = 9
y^2/9-(x+2)^2/9 =1
standard form of parabola, (x-h)^2/a^2-(y-k)^2/b^2 = 1 (opens horizontally)
or (y-k)^2/a^2-(x-h)^2/b^2 = 1 (opens vertically)
this is a hyperbola opening in the vertical direction
Other parameters:
a^2=9
a=3
b^2=9
b=3
c=sqrt a^2+b^2=sqrt 18 =4.24
ctr((-2,0)
foci (-2,±c)
foci (-2,±4.24)
vertex (-2,±a)
vertex (-2,±3)
asymptotes y=±(a/b)x=(3/3)x=x
sorry I don't have ways to draw the hyperbola here, but you should be able to do it with the above information.
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