Question 999213
to write the equation of the ellipse {{{36x^2 +9y^2 +360x-54y+657=0}}}  in standard form {{{(x-h)^2/a^2 +(y-k)^2/b^ 2 =1}}}, complete the squares, group all {{{x}}} and all {{{y}}} together


{{{(36x^2+360x+b^2)-b^2+(9y^2-54y+b^2)-b^2+657=0}}}...factor out  {{{36}}} from first group and {{{9}}} from second group


{{{36(x^2+10x+b^2)-36b^2+9(y^2-6y+b^2)-9b^2+657=0}}}....recall: {{{(a-b)^2=a^2-2ab+b^2}}}

compare to {{{36(x^2+10x+b^2)}}} and you see that {{{a=1}}} and 

{{{2ab=10}}};solve for {{{b}}}
{{{2*1*b=10}}}=>{{{b=5}}}

in {{{(y^2-6y+b^2)}}}=>{{{2ab=6}}}=>{{{b=3}}}


so, we have


{{{36(x^2+10x+5^2)-36*5^2+9(y^2-6y+3^2)-9*3^2+657=0}}}


{{{36(x+5)^2-36*25+9(y-3)^2-9*9+657=0}}}


{{{36(x+5)^2-900+9(y-3)^2-81+657=0}}}


{{{36(x+5)^2+9(y-3)^2-981+657=0}}}


{{{36(x+5)^2+9(y-3)^2-324=0}}}


{{{36(x+5)^2+9(y-3)^2=324}}}....both sides divide by {{{324}}}


{{{36(x+5)^2/324+9(y-3)^2/324=324/324}}}


{{{36(x+5)^2/324+9(y-3)^2/324=1}}}....simplify


{{{cross(36)(x+5)^2/cross(324)9+cross(9)(y-3)^2/cross(324)36=1}}}


{{{(x+5)^2/9+(y-3)^2/36=1}}}


=> {{{h=-5}}}, {{{k=3}}}, {{{a=3}}}, {{{b=6}}}