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=(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