Question 77990: I can't figure out this hyperbola: (((3x^2 + y^2 +18x - 2y + 4=0))). This is what I've tried so far: 3x^2/3 + y^2 + 18x/3 - 2y +4=0
x^2 +y^2 + 6x - 2y =-4
x^2/-4 + y^2/-4 + -3x/2 + y/2 = 1)))
please help me Found 2 solutions by stanbon, scott8148:Answer by stanbon(75887) (Show Source):
You can put this solution on YOUR website! 3x^2 + y^2 +18x - 2y + 4=0
Complete the square for the x-terms and for the y-terms separately.
---------
3(x^2+6x+9) + y^2-2y+1) = -4+27+1
3(x+3)^2 + (y-1)^2 = 24
Divide thru by 24 to get the form of a hyperbola.
[(x+3)^2]/8 + [(y-1)^2]/24 = 1
=========
Cheers,
Stan H.
You can put this solution on YOUR website! since the squared terms are both positive, this is an ellipse rather than a hyperbola...start by regrouping terms
3(x^2+6x+___)+y^2-2y+___=-4...now complete the squares 3(x^2+6x+9)+y^2-2y+1=-4+27+1...so 3(x+3)^2+(y-1)^2=24
dividing by 24 gives
an ellipse centered at (-3,1); with the foci at (-3,-3) and (-3,5)
(