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I've been looking through some questions on this site, and I found this question:
Parabola is an ellipse, but with one focal point at infinity

The top reply has 111 upvotes, and shows a visual to accompany the explanation.
However, I have some doubts that I need to clarify. I do not have 50 points to comment on the post, and I cannot message the person who provided that answer, so I'm hoping that making a post will work.

In the answer with 111 upvotes, I can understand the algebra and the steps, but I have something I'm confused about:

In the visual, why does the (semi)minor axis of the ellipse change? Shouldn't the semiminor axis of the ellipse be independent of the changing eccentricity/semimajor axis/distance-from-focus-to-centre?

helpme
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2 Answers2

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In an ellipse, you have several things that are related to each other:

  • The location of one focus
  • The location of the other focus
  • The distance from one focus to the center
  • The distance from one focus to the nearest vertex (one end of the major axis)
  • The semimajor axis
  • The semiminor axis
  • The eccentricity

and a bunch of other things.

So you certainly can keep the semimajor axis constant and move the foci farther from the center and closer to the vertices of the ellipse. But that is not what is being done in the other answer.

In that answer, the location of one focus is fixed and so is the distance from the focus to the nearest vertex (hence the location of that vertex is fixed as well). Now we change the eccentricity. In order to keep the location of one focus and one vertex fixed, the other focus has to move, and so does the center. Since the distance between the focus and the center changed, but the distance from the focus to the nearest vertex did not, the sum of those two distances (which is the semimajor axis) changed. Through the other relationships among the parts of the ellipse, the semiminor axis changed also.

It's all a matter of which family of ellipses you want to study.

Here are some of the specific measurements of the ellipse and the relationships among them. Let \begin{align} a &= \text{length of semimajor axis},\\ b &= \text{length of semiminor axis},\\ c &= \text{distance from center to focus},\\ e &= \text{eccentricity},\\ \ell &= \text{length of semilatus rectum},\\ p &= \text{distance from focus to the nearest vertex}, \end{align} as in this figure adapated from https://en.wikipedia.org/wiki/File:Ellipse-param.svg:

figure of ellipse with dimensions labeled

(I relabeled parts of the figure to match the equations above, which follow some common conventions for labeling parts of an ellipse or parabola.)

The eccentricity is not labeled, but is it given by the formula $$ e = \frac ca. \tag1 $$

Other relationships among the parameters are \begin{align} a^2 &= b^2 + c^2, \tag2\\ \ell &= \frac {b^2}{a}, \tag3\\ p &= a - c. \tag4 \end{align}

From $(1)$, we get $c = ae,$ so $$ p = a - ae = a(1 - e). $$ Therefore if we hold $p$ constant but allow $e$ to vary, $$ a = \frac{p}{1 - e}$$ and $$ c = ae = p\frac{e}{1 - e}, $$ so $a$ and $c$ both go to infinity as $e$ approaches $1.$

From $(2)$, we have $$b^2 = a^2 - c^2 = (a-c)(a+c) = p(a+ae) = p^2\frac{1+e}{1 - e},\tag5$$ which implies that as $e$ goes to $1,$ $b^2$ goes to infinity, and therefore so does $b$. And that's how the semimajor axis depends on $e$ when you hold $p$ constant.

Plugging $(5)$ into $(3)$, $$\ell = \frac {p(a+c)}{a} = p\left(1 + \frac ca\right) = p(1+e),$$ so as $e$ approaches $1,$ $\ell$ approaches $2p,$ which is the semilatus rectum of a parabola where the distance from the focus to the vertex is $p.$

David K
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    Hmm... how do I find out that the semiminor axis (b) changed as a result of changing semimajor axis (a) and eccentricity (e)? I can recall the relationship between a, b and e to be b^2=a^2(1-e^2). a approaches infinity, and e approaches 1. How exactly does the semiminor axis (b) change since (1-e^2) will become 0 and a^2 will become infinite? – helpme Sep 07 '19 at 21:45
  • Thank you for the detailed response!! It's interesting to see l approaching 2p, which matches up with the visual. One thing I would like to point out is step (5) having p(a+ae)=p((1+e)/(1-e)). I think it should be p^2 ((1+e)/(1-e)) instead. – helpme Sep 10 '19 at 18:32
  • You are absolutely right about the $p^2.$ Fixed! – David K Sep 10 '19 at 18:53
  • Similarly to the considerations from user "helpme", if $p = a(1 - e)$, it is an indeterminate form $\infty \cdot 0$. So, I wonder if and why it is correct to state that $$b^2 = p^2 \frac{1 + e}{1 - e}$$ is infinite, if the value of $p$ is not clearly determined. This is also considered in my question about $b$. – BowPark Sep 23 '19 at 11:13
  • @BowPark In every case $p$ is defined as the distance between a focus and the nearest vertex. Those two points are well defined in all ellipses and parabolas. So "if the value of $p$ is not clearly determined" makes as much sense to me as "if $1 = 0.$" Perhaps you are confused by the fact that it is possible to construct an indeterminate form in an expression for $p.$ But "indeterminate form" does not mean "not determined". It just means that there is a certain set of "determinate" forms that does not cover this particular case. – David K Sep 23 '19 at 11:52
  • @DavidK I wrote in a misleading way. I am sure that this indeterminate form has a value. According to your definition, in a parabola $p$ has a measurable and finite value. Then, should we say that the expression $p = a(1 - e)$ is not appropriate for a parabola? This is what confuses me: $p$ is finite, but it seems that it is impossible to obtain it from $p = a(1 - e)$ (because of the indeterminate form). – BowPark Sep 27 '19 at 16:33
  • Well of course if you start with the assertion that $p$ is a known finite value, how can it possibly be said to be “indeterminate”? The idea of an indeterminate form is really only applicable when you don’t already know the value of the expression. When you have something of the form $p=xy$ where $p$ is known, finite, and non-zero, then the inference is that $x\to\infty$ as $y\to0$ because the product is the known quantity $p.$ That is, we infer something about one of the things on the right-hand side from the other two things in the equation. – David K Sep 28 '19 at 02:47
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Fixing the minor axis is problematic for visualizing the phenomenon at hand, since an ellipse's minor axis is finite, while a parabola's is not. @robjohn's animation, duplicated here,

enter image description here

fixes an element that remains finite throughout: the focus-to-vertex distance.

Another good element to fix is the latus rectum (the focal chord perpendicular to the major axis):

enter image description here

As with @robjohn's animation, one focus remains at the origin while the other scoots off to infinity.

A nice thing about this family is that it's generated by a simple polar equation $$r = \frac{\ell}{1-e\sin\theta}$$ (with $\ell$ the semi-latus rectum) where the animation varies eccentricity $e$ from $0$ (the circle) to $1$ (the parabola). (Taking $e$ beyond $1$ brings hyperbolas into the picture. In that context, one might well convince oneself that the ellipse's focus didn't just go off to infinity to give the parabola, it came back from the other side of the universe to give the hyperbolas.)

Yet another way to visualize the phenomenon is with Dandelin spheres. I'll leave that investigation to the reader.

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