Parabola intersection at line of infinity Does a parabola eventually form a sort of ellipse when stretched to infinity along its axis? I am asking because I am trying to intuitively understand the following picture and the fact that the line at infinity is a tangent of parabola: 

 A: Elipse has $2$ focuses. What would fouces be in that case? 
If we imagine point $F$ in infinity along that axis and $G$ is focus of parabola then for every point $P$ on parabola we have $$PF+PG = \infty $$
Here is an explanation for the tangent. Let $\infty _{\ell}$ means point in infinity determined by line $\ell$.

For every $P$ on a parabola, perpendicular bisector (= MP) for $P'G$ is a tangent on parabola at $P$. So as $P'\to \infty_f $ also $P\to \infty_{axis}$ and $M\to \infty_f$. Now if $P'=\infty _f$ then $MP = \infty_f\infty_{axis}$ = line in infinity.
A: When talking about weird things like „line at infinity“ it is best to be clear on the spaces one is using.

Edit in a comment to the accepted answer the op has mentioned that the context is the projective plane. What follows is therefore irrelevant. However I feel like the viewpoint I present is interesting enough to not delete my answer. I apologize for potential confusion that arose.

So I guess you are talking about the extension of the polynomial $p:\mathbb R \rightarrow \mathbb R, x \mapsto x^2$ to the one point compactification $\mathbb R_\infty=\overline{\mathbb R}/\{\pm\infty\} \cong \mathbb S^1$ by letting $p(\infty) = \infty$. The line at infinity is the constant map $\ell:\mathbb R_\infty \rightarrow \mathbb R_\infty, x\mapsto\infty$.
Now think of the graph of both $p$ and $\ell$ in $\mathbb R_\infty \times \mathbb R_\infty \cong \mathbb S^1\times\mathbb S^1$, which is the torus. It is hard to describe but not too hard to see that the graph of $p$ forms a closed path around the torus meeting the graph of $\ell$ only once. So in some sense, the graph of $p$ indeed forms a cycle. 
Writing $c_0$ for the constant function evaluating to 0 the attached image should illustrate what I mean.

