Question 894166
I will only take this part of the way and you can finish it:


{{{4x^2-4x+4y^2-8y=1}}}
{{{4(x^2-x)+4(y^2-2y)=1}}}


You need to Complete The Squares.
The term for x is {{{(-1/2)^2=1/4}}}
The term for y is {{{(-2/2)^2=1}}}


{{{4(x^2-x+1/4)+4(y^2-2y+1)=1+4(1/4)+4*1}}}, note that is equivalent to adding the two square terms to both sides of the equation.


{{{4(x-1/2)^2+4(y-1)^2=1+1+4}}}
.
.
Now, you continue and finish.