Question 1045637
the standard form of the equation {{{x^2+3y^2+4x+6y+1=0}}} will be:

{{{x^2+3y^2+4x+6y+1=0}}}......group like terms

{{{(x^2+4x)+(3y^2+6y)+1=0}}}

{{{(x^2+4x)+3(y^2+2y)+1=0}}}......complete squares


{{{(x^2+4x+b^2)-b^2+3(y^2+2y+b^2)-3b^2+1=0}}}.....recall, {{{(a+b)^2=a^2+2ab+b^2}}} and in your case for {{{(x^2+4x+b^2)}}} we have {{{a=1}}} and {{{2ab=4}}}, so we can find {{{b}}}

{{{2*1*b=4}}}->{{{b=4/2}}}->{{{b=2}}}

for {{{3(y^2+2y+b^2)}}} we have {{{a=1}}} and {{{2ab=2}}}, so we can find {{{b}}}

{{{2*1*b=2}}}->{{{2b=2}}}->{{{b=1}}}

then we have

{{{(x^2+4x+2^2)-2^2+3(y^2+2y+1^2)-3*1^2+1=0}}}

{{{(x+2)^2-4+3(y+1)^2-3+1=0}}}

{{{(x+2)^2+3(y+1)^2-7+1=0}}}

{{{(x+2)^2+3(y+1)^2-6=0}}}

{{{(x+2)^2+3(y+1)^2=6}}}..............both sides divide by {{{6}}}

{{{(x+2)^2/6+3(y+1)^2/6=6/6}}}

{{{(x+2)^2/6+cross(3)(y+1)^2/cross(6)2=1}}}


{{{(x+2)^2/6+(y+1)^2/2=1}}}-> you have an ellipse and this is the standard form of the equation