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A parabola can be regarded as an ellipse with a focus to infinity.

Its semi-major axis $a$ has an infinite length, and this is obvious. But what are the implications of a semi-major axis $a \to \infty$ and an eccentricity $e = 1$ on the semi-minor axis $b$? Is it infinite as well?

My attempt

Assuming that it is a limit-case of an ellipse:

$$e = \sqrt{1 - \frac{b^2}{a^2}}$$

If $b$ is fixed and finite, as $a \to \infty$, $e \to 1$ as in a parabola. From the above definition,

$$b^2 = a^2 (1 - e^2)$$

but this is an indeterminate form $\infty \cdot 0$, being $a^2 \to \infty$ and $1 - e^2 \to 0$. Also, from the relation $a^2 - c^2 = b^2$ (where $c$ is the distance of each focus from the origin), another indeterminate form $\infty - \infty$ is obtained.

Also, this answer shows a procedure about $b$, but, as pointed out in the comments, $p = a(1 - e)$ is still an indeterminate form $\infty \cdot 0$.

This question is related to a previous one.

BowPark
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  • I’m thinking about it, but consider that the axes bisect each other. This does not make sense here as there isn’t really a “midpoint” of the major axis. And, after all, $b$ is in an indeterminate form and $e$ and $a$ are related by $b$ itself. – Certainly not a dog Sep 23 '19 at 11:32
  • Related: "A parabola is an ellipse with one focus at infinity". – Blue Sep 23 '19 at 11:42
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    You have found some indeterminate forms in a set of limits. So what? Do you claim that L'Hopital's Rule is invalid because it supposedly finds the limit of an indeterminate form? Keep in mind that the whole point of a "limit case" is that there are typically formulas that are valid as you approach the limit but not at the limit. So finding a few formulas like that can explain why we bothered with the notion of a "limit case" in the first place, but if you want to make more of it than that I think you need to explain your concerns further. – David K Sep 23 '19 at 11:50
  • A minor axis of finite length would mean that the parabola has asymptotes parallel to the major axis. Which it doesn't have. –  Sep 23 '19 at 12:23
  • @DavidK I'm not sure I got your point in the comment. The paragraph "My attempt" tries to highlight some contradictions in my procedure, not in any procedure. I do not want to show the problems arising with indeterminate forms: I would like instead to analytically - and coherently - show if $b$ is infinite. – BowPark Sep 27 '19 at 16:23
  • @YvesDaoust This is a good starting point for a proof. But is it enough? Sorry, I am not familiar with such procedures. – BowPark Sep 27 '19 at 16:25

1 Answers1

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I think it could be helpful to remember what a comic section is. It is the intersection of an infinite double circular cone with a plane.

Everything we say about conic sections involving a focus, a vertex, a major axis, a minor axis, a directix, a transverse axis, a conjugate axis, a center, or an asymptote is derived from properties of the intersection of a cone and plane under certain sets of circumstances. As far as I can see, none of the definitions of these features is applicable to every conic section, even if we exclude degenerate cases.

If we are working in a projective plane, a parabola actually is an ellipse with one point on the line at infinity. But in an ordinary Euclidean plane I do not think it is strictly correct to say that a parabola “is” an ellipse. I would rather say that a parabola is a limiting case of an ellipse as the eccentricity goes to $1.$

And of course a single ellipse is exactly one set of points with one eccentricity. The eccentricity of that particular set of points does not somehow change. To make the eccentricity go to $1,$ we need an infinite sequence of ellipses with eccentricities that approach $1$ in the limit.

Moreover, a parabola is merely a limiting case of a sequence of ellipses, not the limiting case as the eccentricity goes to $1.$ We have many choices about how to make the eccentricity of a sequence of ellipses go to $1.$ For example, we could hold the center fixed and hold the length of the semiminor axis fixed, but increase the distance between the foci without bound. We would then have a sequence of ellipses all the same width but stretched out longer and longer. What is the limiting case of this sequence? Maybe a pair of straight lines. Anyway, it is not a parabola. But the limit of the semiminor axis is a finite nonzero length.

Another alternative is that we hold one focus and the length of the semimajor axis fixed, and move the other focus away from the fixed one. As the distance between foci approaches the length of the major axis, the eccentricity approaches $1$ and the semiminor axis approaches zero. The limiting figure is a line segment, not a parabola.

But these are weird sequences, especially if you look at what cones you need in order to produce them. In the first example we need to send the apex off toward infinity and shrink the opening angle toward zero. In the second example we need the apex to go onto the intersecting plane in the limit.

But there are other ways to take the eccentricity toward $1$ without getting so weird. One example that is easy to calculate is to fix the location of one focus and the nearest vertex of the ellipse. Then the distance between these points, $p,$ is constant, and the formula $p=a(1-e),$ far from being indeterminate in any way (it has a constant value!) shows that $a$ must go to infinity as $e$ goes to $1.$

The formula $b^2=a^2(1-e^2)$ by itself is not very informative as $a\to\infty$ and $e\to 1,$ but combined with $p=a(1-e)$ it implies that $b^2=a(1+e)p.$ Then holding $p$ constant, as $e\to1$ and $a\to\infty$ the right-hand side of the equation goes to infinity, and therefore so does $b.$

(Do you see what happened there? There was a quantity, $b^2,$ that when expressed as a product of two expressions in a particular way gave us an indeterminate form, but rearranging the terms in the product produced something whose limit can be determined by simple rules.)

So that is what happens to $b$ according to one way of taking the eccentricity of a sequence of ellipses to $1.$ In general, however, notice that no parabola fits between two pairs of parallel lines, so if we have a sequence of ellipses that all have semiminor axes less than some upper bound, there are parts of any parabola that they can never get close to. So no such sequence of ellipses can have a parabola as a limiting case; to approach a parabola, $b$ must go to infinity.

David K
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  • A massive thank you for this thoroughful explanation. Assuming that «a parabola can be regarded as an ellipse with a focus to infinity» I was also struggling to obtain a parabola by simply getting the foci closer to the vertices, as in the "another alternative" you mentioned. Now this is clear, as well as the fact that an expression leading to an indeterminate form can be rearranged to obtain, if it exists, a determinate value. – BowPark Sep 28 '19 at 06:21