Question 1180973

{{{16x^2+25y^2-160x-200y+400=0}}}

to convert the general equation  to standard form, complete squares

{{{(16x^2-160x)+(25y^2-200y)+400=0}}}


{{{16(x^2-10x)+25(y^2-8y)+400=0}}}


{{{16(x^2-10x+b^2)-16b^2+25(y^2-8y+b^2)-25b^2+400=0}}}...for {{{x}}} part {{{b=10/2=5}}}, and for {{{y}}} part {{{b=8/2=4}}}


{{{16(x^2-10x+5^2)-16*5^2+25(y^2-8y+4^2)-25*4^2+400=0}}}


{{{16(x-5)^2-400+25(y-4)^2-400+400=0}}}


{{{16(x-5)^2+25(y-4)^2=400}}}.........both sides divide by {{{400}}}


{{{16(x-5)^2/400+25(y-4)^2/400=400/400}}}


{{{(x-5)^2/25+(y-4)^2/16=1}}}