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

While I was reading about conic sections, I came across the following statement:

A parabola is an ellipse, but with one focal point at infinity.

But it is not clear to me. Can someone explain it clearly?

• See also projective plane/point at infinity. – imallett May 2 '14 at 19:08
• The funny thing is I've always thought of conic sections this way (after I learned about them). A parabola is an ellipse with a focal point at infinity; it is also a hyperbola with a focal point at infinity. To get from an ellipse to a hyperbola, the point wraps around at infinity. This seemed even more logical when I learned about eccentricity. – Justin May 3 '14 at 3:14
• @Ian Mallett: the concept of a focal point fails in projective geometry: there is no distances. Note also that the supposed “focal point” lies on the curve (parabola) in ℝP², so it would fail even if there were distances. – Incnis Mrsi Sep 1 '14 at 11:49
• @IncnisMrsi See e.g. jrh794.wordpress.com/2012/10/22/…. In the compactification of $R^2$, the point at infinity is an actual point, and you can see how the parabola is an ellipse. – imallett Sep 1 '14 at 17:15
• @Ian Mallett: did you actually read what I said? I obviously know what is ℝP² and how the line at infinity is a tangent to a parabola, but there are no focal points in projective geometry, period. – Incnis Mrsi Sep 1 '14 at 19:10

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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• Following on this line-of-thought, an hyperbola is a ellipse which had its focal point "wrap-around" at infinity and then come back from the other side? – NothingsImpossible May 3 '14 at 0:40
• @NothingsImpossible: indeed – robjohn May 3 '14 at 11:36
• With $a \to \infty$, in the last step you simplify the term $y^2 / a$, considering this term as $0$. Why? There are some points still belonging to the ellipse where $y$ approaches infinity, so it may be comparable with $a$. This, in particular, happens at the opposite vertex. Why then is it possible to consider $y^2 / a \to 0$? – BowPark Sep 30 '19 at 11:47
• @BowPark: Of course, an ellipse is never actually a parabola. However, all points on the ellipse within a given distance of the origin (which is what we can "see", that is, plot), do tend to points on the given parabola. For $|(x,y)|\le R$, we have $\frac{y^2}a\le\frac{R^2}a\to0$. – robjohn Sep 30 '19 at 13:49

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.

Think of a pair of cones with vertical axes and identical cone-angles, one facing up and one facing down. Place them so they're touching at one point. I'll call this a "cone" for now.

If you slice this object with a plane perpendicular to the axis, you'll get a circle of some radius, or perhaps a single point, which you could call a circle of radius 0.

If you slice it with a slightly tilted plane, you'll get an ellipse (or a single point). Thus circules and ellipses are both "cross-sections" of a cone, or "conic sections".

If you tilt the slicing plane further, so it's nearly vertical, it'll intersect both cones, resulting in a hyperbola, or, if the plane passes through the cone-point, a pair of intersecting lines. So intersecting lines are kind of a "limit" of hyperbolas.

Go back to ellipses: place the plane to slice an ellipse from the upper half of the cone. Tilt your slicing plane more and more, making the ellipse more and more eccentric. There's a last possible amount of tilt before you start also slicing into the bottom part of the cone. At that tilt, the intersection is no longer an ellipse, but instead a parabola.

So it's reasonable to say that a parabola is a limit of ellipses.

Of course, if you tilt a tiny bit more, you start getting hyperbolas, so it's also reasonable to say a parabola is a limit of hyperbolas.

As other answers show it makes perfectly sense to consider a parabola as limiting member of a family of ellipses. One might also call it a conic section touching the line at infinity, and I'm sure that other visualizations are possible.

But I don't think that you can say that in the limit "one of the foci is at infinity". We have to face the fact that in the limit one of the foci has disappeared once and for all. Foci belong strictly to euclidean geometry and to "finite" ellipses and hyperbolas in the euclidean plane. Already affine mappings destroy their distinguished character, let alone projective transformations coming into play when we talk about points at infinity.