Parabola is an ellipse, but with one focal point at infinity

The equation for an ellipse with a focus at $(0,0)$ and the other at $(0,2ae)$ keeping $a(1-e)=f$ (where $f$ is distance from the vertex to the focus of the ellipse, which ends up being the focal length of the parabola) is
$$\frac{x^2}{a^2(1-e^2)}+\frac{(y-ae)^2}{a^2}=1$$
which is equivalent to
$$\frac{x^2}{f(1+e)}+\frac{y^2-2aey}{a}=f(1+e)$$
If we let $a\to\infty$ (and therefore $e=1-\frac fa\to1$), we get
$$y=\frac{x^2}{4f}-f$$
which is a parabola.

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