Question 77990
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 {{{(((x+3)^2)/8)+(((y-1)^2)/24)=1}}}


an ellipse centered at (-3,1); with the foci at (-3,-3) and (-3,5)