Which conic section does the equation below describe?2x²+2y²-6x+4y+ 1 = 0

Answer :

we have the equation

[tex]2x²+2y²-6x+4y+1=0[/tex]

Group similar terms and move the constant term to the right side

[tex](2x^2-6x)+(2y^2+4y)=-1[/tex]

Factor the leading coefficient on both terms of the left side

[tex]2(x^2-3x)+2(y^2+2y)=-1[/tex]

Complete the square twice

[tex]2(x^2-3x+1.5^2)+2(y^2+2y+1)=-1+4.5+2[/tex]

Rewrite as a perfect square

[tex]2(x-1.5)^2+2(y+1)^2=5.50[/tex]

Divide both sides by 2

[tex]\begin{gathered} (x-1.5)^2+(y+1)^2=\frac{5.50}{2} \\ \\ (x-1.5)^2+(y+1)^2=2.75 \end{gathered}[/tex]

we have the equation of a circle

therefore

The answer is a circle