Question

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

Answer 1

we have the equation

$2x²+2y²-6x+4y+1=0$

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

$(2x^2-6x)+(2y^2+4y)=-1$

Factor the leading coefficient on both terms of the left side

$2(x^2-3x)+2(y^2+2y)=-1$

Complete the square twice

$2(x^2-3x+1.5^2)+2(y^2+2y+1)=-1+4.5+2$

Rewrite as a perfect square

$2(x-1.5)^2+2(y+1)^2=5.50$

Divide both sides by 2

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

we have the equation of a circle

therefore

The answer is a circle