The Ultimate Fate of Conic Sections.

Factual Questions
1

I have been fascinated by science–and particularly math–all my life. I am more of a novice that a professional, when it comes to the subject. But that doesn’t mean I can’t still have fun with it.

Anyways, I learn about conic sections in high school, as I suppose most of us do. The circle, the parabola, the hyperbola and the ellipse all come from planes intersecting this double three-dimensional cone. Why this is so, or how exactly it is done, I don’t know. I never got educated above the two dimensions. Sorry.

But I have seen illustrations, of how these functions come from intersection the double cone.

The parabola comes from intersecting the cone, parallel to its side. And I have always wondered. If taken to zero, would the parabola become a line? I know the function at the origin is Y=AX^2, where A is the constant, that tells you have broad and blunt the parabola becomes. I guess I am asking what happens when A equals zero. But that is easy enough. Then Y=0 and that indeed is a line. That was easy enough. But unfortunately, it becomes more complicate after that.

The one that really fascinates me is the circle. The graphing equation for the circle is (X-A)^2+(Y-B)^2=R, where R is the radius. But what if R equals zero (0)? Would that, as I imagine, be a point? And how on earth do you diagram a point? Also, to turn it into a function (which is often used in calculus, I believe), you have to turn the equation into a semi-circle. I don’t know if you can graph a semi-circle into the equation I just gave. But I do know, a semi-circle at the origin, would have the equation F(X) equals the square root of the quantity of (R-X^2). But when R (the radius) equals zero, you get the square root of negative X. You can’t take a square root of a negative number. What am I missing here?

Anyways, to move on, we have the hyperbola. For some reason, we never covered the hyperbola in hs math class. But from what I understand, it is intersection of the double cone, with a perfection vertical line. And if it hits the dead center of the cone, it would two intersection lines. Wouldn’t it? As I said, I know very little about hyperbolas. So you will just have to help me out more with that one.

What is the solution to the problems I have submitted? And again, what am I missing here (especially with a circle “degenerating” into a point)?

I look forward to your replies:). As I said, I don’t know about the rest of you. But I find this topic fascinating, even if it only involves basic high school math;).

:):):slight_smile:

2

I’ve only got this.

I, too, remember learning conic sections in high school. When we did the parabola, my ears perked up a bit, having just read a cheesy sci-fi story in which perfect reflectors shaped into perfect parabolas played a part in the defense of Earth. In this discussion, parabolas were defined as the set of all points equidistant from a point and a line. As both points and (infinite) lines don’t really have any *size *to speak of, I came to realize that all parabolas are similar the same way triangles with equal angles are- that is they can all be adjusted by scaling to match each other.

This is NOT true of ellipses and it does not seem to me to be true of hyperbolas.

But I came back full circle to… circles. Circles are the set of all points equidistant from a central point. And all circles are similar- that is the same after scaling. Well, duh, they look the same.

The thing is, although I had long known how to *use *pi, I’d never before thought much about why there should be a constant ratio between diameter and circumference. I kind of felt clever and like an idjit at the same time.

3

Yep! If the parabola goes through the intersection point – the middle of the hourglass – where the two cones join, then it degenerates to an “X” shape, going right up each cone, to left and right, forever.

If you move a circle up to that point, it degenerates to a point.

If you move a hyperbola to meet that point, it degenerates into a “V.” (Oops, I think. I’m not actually sure on that. I’m also not sure what you get if you move an ellipse up, so part of it is above the joining point and part of it is below. Some sort of figure eight? I’m not succeeding in visualizing it. Need a new app for my brain.)

(I saw a cartoon once of a lump of stuff, drooling and leering and making cat-calls, and inviting other lumps of stuff to engage in kinky sex with it. The caption was “degenerate matter.”)

ETA: The “yep” was to the OP re parabolas, not to Blue Blistering Barnacle for being an idjit! Also, as a Tintin fan, I love your user name!

4

/hijack/

Yay! Tintin!

/hijack/

5

A parabola degenerates to a single line, not to two of them, and a hyperbola degenerates to an X shape, not a V. But other than that, yes, those are the correct degenerate forms of those figures. Similarly, you can have (for instance) a degenerate rectangle, which is just a line segment or even a point.

Are degenerate geometric figures interesting, or worthy of study? Well, that depends on the context. For instance, we say that orbits have the shape of conic sections, and an object just falling straight down in a line is a valid orbit.

And one minor addendum to the OP:

The plane does not need to be perfectly vertical, though that’s usually the way it’s depicted in geometry books. Any angle steeper than the sides of the cone will work. Also, not all hyperbolae are similar to each other: When you zoom out on them, even the non-degenerate ones look an awful lot like an X, and that X can have any angle at all. Hyperbolae with different angles are not similar to each other.

6

This seems like a [del] good[/del] perfect place to ask a question that has been bugging me for about a half-century. I think the OP is totally on track, and if anyone thinks I’m hijacking the thread, then I apologize.

An ellipse is symmetrical along both axes. I have no problem with that. After all, an ellipse is just a wide circle. Or a tall circle, if you prefer. Or, if you want to get pedantic, a circle is actually a special sort of ellipse. Whatever.

My problem is when I look at the conic section that forms the ellipse. It doesn’t make sense to me that it is symmetrical on both axes. If I put my hand on a cone close to the vertex, I can easily tell that the curves are very tight. Far from the vertex, the curves are wider. And if I go VERY far from the vertex, the cone seems almost flat. Got the picture?

So the intuitive thing to me, is that this conic should be symmetrical along the long axis, but not along the short one. More specifically, it should be more pointed at the end near the cone’s vertex, and more rounded at the far end. It should look similar to an egg that was cut along the long axis.

I know that I am wrong. I know that the conic section called an ellipse IS symmetrical on both axes. My question is WHY?

Example: Let’s take a cone of medium to narrow width, say 20 to 50 degrees. Let’s say that one end of the intersecting plane is only one millimeter from the vertex, and the other end is many meters away. Intuitively, I imagine the resulting ellipse will have a very tight curve - almost sharp and pointed - near the vertex. And the other end will be far less pointy. (I know that the portion of the cone that is cut off will be very thin, but I’m asking about the shape of the perimeter.) But that’s my imagination, and I’m told that the reality is different.

I’ve never seen any actual cones and ellipses that would demonstrate how wrong I am. The vast majority of ellipse illustrations show the long axis only a bit longer than the short, and this makes it hard to tell whether it is egg-shaped or symmetrical.

Does anyone know what I’m trying to describe? Are the any websites with good illustrations?

advTHANKSance

7

Well, you diagram a point by drawing a point. I’m not sure I understand your problem with this. The graph of the function (x-a)[sup]2[/sup] + (y-b)[sup]2[/sup] = 0 is the point at (a,b).

The equation of the semicircle above the x axis for your equation would be
y = b + sqrt(R[sup]2[/sup] - (x-a)[sup]2[/sup])

As you can see from the above, the value under the square root is R[sup]2[/sup]-(x-a)[sup]2[/sup]. When the value of (x-a)[sup]2[/sup] is larger than R[sup]2[/sup], so that the quantity becomes negative, the function ceases to have a real y value. This just means that the graph of the semicircle is confined to the region between a-R and b+R, as you would expect. The graph doesn’t reach to x values that are at a distance more than R from the center at a.

8

I don’t follow that exaclty, but what you are missing is that a circle is only defined for values where x is part of the circle. When R=0, F(x) is only defined over the range x=0.

The part of your question that I don’t follow is the part where I would explain that the square root of negative X does have a solution: a complex number including an imaginary part. And that these are conic “solutions” where the plane does not intersect the cone. And that the set of complex “solutions” are equally important and interesting.

9

The radius of curvature at any point where the cone is cut by a plane is going to depend on the angle of the plane and the azimuth of the vector normal to the plane (orthogonal to the long and short axes of the ellipse, and will be the same at opposing points (at 180 degrees to each other) around the ellipse. The curvature of the cone is of course constant, even as the radius increases (by definition; a cone is essentially a ramp wrapped around an axis), so the profile will be the same regardless of where the plane is with regard to the origin; the only thing that changes is the size of the ellipse. Here is a reasonably good depiction of the planar cut across a cone creating an ellipse.

Stranger

10

Should read “between a-R and a+R”.

11

Le sigh! I miss being correct by small margins in this way…an awful lot of the time. Thank you for the correction!

I remember degenerate triangles – the lengths of side a and side b add up to the length of side c – and how they were fun traps and trick-questions in high school geometry classes.

12

I thought this thread was going to be about newspapers getting rid of the funnies.

13

Thanks. It is indeed a good example of how to make an ellipse, but the cut is so close to perpendicular that it does not demonstrate my point. But its suggestions to use Excel’s “chart” feature give me some tools for further research. Thanks.

14

The Greeks came up with these things, right? Does anybody know how? They didn’t have analytical geometry yet, and I’m not sure how one creates or manipulates conic sections with compass and straightedge. What did the study of conic sections look like to a mathematician, circa 300 BC, who wasn’t familiar with decimal numbers or equations, let alone calculus?

15

Keeve, it might help to think of it like this: You’re right that the cone forms smaller circles near the vertex, and larger circles further away. But the angle that the plane makes with the two sides is different.

If you take a cylinder, and cut it with a plane, you’ll also make an ellipse. If the plane is close to perpendicular to the cylinder, then the ellipse you get will be close to a circle, but if the plane is tilted a lot, you’ll get a very long ellipse, very pointy at the ends.

Well, when you cut a cone with a plane, on one side (where it’s closer to the vertex) the plane will be close to perpendicular to the edge of the cone, but on the other side (further from the vertex), it’ll be tilted a lot more. So one side of the ellipse is pointy because it’s from where the circles are small, and the other side of the ellipse is pointy because of the angle of the plane. And if you do the math, it works out that both effects have equal strength, and so the ellipse ends up being equally pointy at both ends.

16

Doubling cubes. On flat paper.

The question is a standard one in histories of mathematics. As to the conceptual bases, you’ve got a different history to follow. For example, from the Stanford Encyclopedia of Philosophy, “Epistemology of Geometry”:
…In its synthetic form the successes of projective geometry were largely confined to the simplification it brought to the study of conics—all non-degenerate conics (the circle, ellipse, parabola, and hyperbola) are projectively equivalent. In its algebraic form projective geometry proved itself almost essential in the study of plane algebraic curves of any degree, and, extended to projective spaces of higher dimensions, to the study of algebraic surfaces. All this contributed to the central importance attributed to a non-metrical geometry based on little more than the concept of the straight line and on the incidence properties of lines and planes.

As to the history of working it out, this undergraduate term paper by Ken Schmarge is as good as any to start.

(The paper is from 1999. Wonder how Schmarge wound up.)

17

If this entirely unproven search is accurate, he’s a golf coach and run over 20 triathlons and duathlons, and…wait for it…has taught math at Ramsey H.S. for 16 years. YES!

Or else he’s somebody else.

18

don’t forget the definition of an ellipse is also that the sum of the distances from any point on the curve to the two foci is a constant. in practical terms, you can take a loop of string and put it around two pegs at the foci of the ellipse, and stretch it out with a pencil. keep the loop taut and it will describe an ellipse. And, of course, the circle is the special case where the two focii are actually at the same point. The Greeks would have no problem drawing ellipses.

19

Somehow the lack of proper orthography in postings about mathematics is particularly grating.

20

Don’t forget that the ellipses (and conic sections in general) don’t need two foci to be defined. The common definition of a parabola uses a focus and a directrix to describe its shape. The same method can be used for ellipses and hyperbolae.

As to the OP:

Actually, any plane that intersects the cone (both top and bottom if you will) regardless of whether is it perfectly vertical or not describes an hyperbola. Being parallel to the axis of the cone is not a necessary condition.