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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Imagine an ellipse made of reflective material. Light rays emanating from one focus and reflecting off the ellipse will all be reflected toward the other focus. (This, applied to sound waves rather than light rays, is the principle behind whispering galleries.) Now imagine instead a parabola made of reflective material. Light rays emanating from its focus and reflecting off the parabola will all be reflected in the direction parallel to the axis of the parabola. (An approximation to this seems to be involved in automobile headlights.) So, if you think, as in projective geometry, of parallel lines as "meeting at infinity", then the point at infinity on a parabola's axis plays the same role as the other focus of an ellipse.