Question 970906
      

{{{x^2-4y^2-6x-16y+29=0}}}

{{{(x^2-6x)-(4y^2+16y)+29=0}}}.....complete square

{{{(x^2-6x+b^2)-b^2-4(y^2+4y+b^2)-b^2+29=0}}}...recall the rule {{{(a-b)^2=a^2-2ab+b^2}}}

in your case (for {{{x}}}) {{{a=1}}}, and {{{2ab=6}}}=>{{{2*1b=6}}}=>{{{b=3}}}

and (for {{{x}}}) {{{a=1}}}, and {{{2ab=4}}}=>{{{2*1b=4}}}=>{{{b=2}}}

so, we have

{{{(x^2-6x+3^2)-3^2-4(y^2+4y+2^2)-(-4)*2^2+29=0}}}

{{{(x-3)^2-9-4(y+2)^2+16+29=0}}}

{{{(x-3)^2-4(y+2)^2+16+29-9=0}}}

{{{(x-3)^2-4 (y+2)^2+36 = 0}}}........both sides divide by {{{-36}}}

{{{(x-3)^2/-36-4 (y+2)^2/-36+36/-36 = 0}}}

{{{-(x-3)^2/36+ (y+2)^2/9-1 = 0}}}

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


{{{ graph( 600, 600, -10, 10, -10, 10, sqrt(((x-3)^2/36+1)9)-2,-sqrt(((x-3)^2/36+1)9)-2) }}}