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If the affinization of a projective variety is irreducible, but not dense in its projective closure (which are the easy cases I've come across in exercises), is the projectivisation still irreducible?

I'm looking at the classic 'saddle' variety $$X = Z \left(\frac{x^2}{p^2} - \frac{y^2}{q^2} - 2zt\right) $$ and trying to show that it's smooth (easy) and irreducible (harder).

I can get as far as showing that the affinisation $X \cap \mathbb{A}^3_{t \neq 0}$ is irreducible (either going by contradiction or considering it over $k(y)$.

I'm left, obviously, with two 'lines at infinity' $\frac{x}{p} = \pm \frac{y}{p}$ in $\mathbb{P}^2$ meeting at $(0:0:1:0)$ and I'm groping around for a topological argument that says that the projectivisation is the closure.

I guess I need to somehow argue that either $X$ being reducible is a contradiction, or show that the two lines aren't a separate irreducible component of $X$ but I'm struggling to frame the topological argument rigorously. Any help gratefully recieved...

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