Question 552369: Can you please show me the graph of 3x^2-2y^2+12x+4y+4=0 and the coordinates of the center and foci please! Answer by lwsshak3(11628) (Show Source):
You can put this solution on YOUR website! Graph of 3x^2-2y^2+12x+4y+4=0
and the coordinates of the center and foci please!
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3x^2-2y^2+12x+4y+4=0
complete the square
3(x^2+4x+4)-2(y^2-2y+1)=-4+12-2
3(x+2)^2-2(y-1)^2=6
divide by 6
(x+2)^2/2-(y-1)^2/3=1
This is an equation of a hyperbola with horizontal transverse axis.
Its standard form: (x-h)^2/a^2-(y-k)^2/b^2=1, (h,k) being the (x,y) coordinates of the center.
For given equation:
Center: (-2,1)
a^2=2
b^2=3
Foci:
c^2=a^2+b^2=2+3=5
c=√5≈2.24
Foci=(-2±c,1)=(-2±√5,1)=(-2±2.24,1)=(-4.24,1) and (.24,1)
see graph below as a visual check on the above:
y=±((3(x+2)^2/2)-3)^.5+1
(