Question 519133: Describe the graph of y^2+4y-3x^2+6x=11. What is the focus?
Answer by lwsshak3(11628)   (Show Source):
You can put this solution on YOUR website! Describe the graph of y^2+4y-3x^2+6x=11. What is the focus?
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y^2+4y-3x^2+6x=11
complete the square
-3(x^2-2x+1)+(y^2+4y+4)=11-3+4=12
-3(x-1)^2+(y+2)^2=12
(y+2)^2-3(x-1)^2=12
(y+2)^2/12-(x-1)^2/4=1
This is an equation of a hyperbola with vertical transverse axis of the standard form:
(y-k)^2/a^2-(x-h)^2/b^2=1, with (h,k) being the (x,y) coordinates of the center.
For given equation:
center: (1,-2)
a^2=12
b^2=4
c^2=a^2+b^2=12+4=16
c=√16=4
Foci: (1,-2±c,)=(1,-2±4)=(1,2) and (1,-6)
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