Question 884823
{{{y^2-25x^2+4y+50x-46=0}}}
{{{y^2-25x^2+4y+50x=46}}}
{{{y^2+4y-25x^2+50x=46}}}
Recognizing that {{{y^2+4y}}} is part of =(y+2)^2}}} ,
we add {{{4}}} to both sides of the equal sign to complete one square on the left side:
{{{y^2+4y+4-25x^2+50x=46+4}}}
Replacing {{{(y+2)^2}}} for {{{y^2+4y+4}}} (on the left of the equal sign),
taking out {{{(-25)}}} as a common factor (also on the left),
and just performing indicated operations on the right side:
{{{(y+2)^2-25(x^2-2x)=50}}}
Recognizing that {{{x^2-2x}}} is part of {{{x^2-2x+1=(x-1)^2}}} ,
we add {{{(-25)*1=-25}}} to both sides of the equation to complete another square:
{{{(y+2)^2-25(x^2-2x)-25*1=50-25}}}
{{{(y+2)^2-25(x^2-2x+1)=25}}}
{{{(y+2)^2-25(x-1)^2=25}}}
Dividing both sides by {{{25=5^2}}} we get:
{{{(y+2)^2/25-25(x-1)^2/25=25/25}}}
{{{(y+2)^2/25-(x-1)^2/1=1}}}
{{{(y+2)^2/5^2-(x-1)^2/1^2=1}}}