Points at infinity of $x^2-6xy+9y^2-3z+1=0$ (in the projective space $\mathbb{P^3}$) (Have I done it well?)

Question

We have to calculate the points at infinity of the following curve:

$x^2-6xy+9y^2-3z+1=0$ (in the projective space $\mathbb{P^3}$)

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I know how are done this type of exercise but in this one exactly, I've got a doubt.

I've done the following:

$D=${$(x,y,z)\in\mathbb{R^3}|x^2-6xy+9y^2-3z+1=0$}

To have a homogeneous equation:

$\mathbb{D}=${$(x:y:z:t)\in\mathbb{P^3}|x^2-6xy+9y^2-3zt+t^2=0$}

So the points at infinity:

$I.P.(\mathbb{D})=${$(x:y:z:0)\in{\mathbb{P^3}|(x:y:z:0)\in{\mathbb{D}}}$}={$(x:y:z:0)\in{\mathbb{P}^3}|x^2-6xy+9y^2=0$}

So, we can differentiate the following cases:

• If $\space $ $y=0 \to x=0 \to(x:y:z:0)=(0:0:z:0)$
  

  And if $z=0$, $(0:0:z:0)=(0:0:0:0)$ and this is not possible because $(0:0:0:0)$ is not a projective point.
  

  But if $z\neq0$, we have $(0:0:z:0)=(0:0:1:0)$
  

• If $y\neq 0\to x=3y$ so we have $(x:y:z:0)=(3y:y:z:0)$ But what does this mean? That it has $\infty$ points at infinity? Which are like $(3:1:z:0)$ ???

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Have I done the exercise well? Or, if not, I would like to know what's wrong and how can I do it in a correct way.

Answer 1

You're doing well: first consider the associated homogeneous polynomial $$ x^2-6xy+9y^2-3zt+t^2=0 $$ and then set $t=0$, which yields $x^2-6xy+9y^2=0$, so $x=3y$. There's no condition on $z$, so you get a whole line, namely $(3:1:k:0)$ and the point $(0:0:1:0)$.