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Today I was discussing circles, ellipses, hyperbolas, and parabolas in my precalculus class. We did the usual: completing the square, finding the center and radius (radii), etc. etc. But I like to also go a little bit deeper on this topic:

Algebraically, all these shapes are related by the fact their implicit equations are quadratic. I like to show this relation geometrically by first drawing a circle, and then "stretching" it to make an ellipse, and then "stretching" it even further to make a parabola (point goes to infinity), and then "stretching" it even further to get a hyperbola. I do this in the projective plane (I draw a big orange circle around the axes, which represents the line at infinity). The students really like it.

But one of my more clever students asked me what is happening to the equation as I'm doing this stretching. So if I start with the equation of the unit circle $$ x^2+y^2=1,$$ and then I do some stretching in the vertical direction, $$ x^2+\left(\frac{y}{b}\right)^2=1,$$ (so here stretching by a factor of $b$) and I let $b$ get really really big, I should expect to get the equation $$ x^2=1,$$ which is just two vertical lines (and that is what I get geometrically!). The students seem to intuitively understand this, and everyone is happy.

If I want to do the parabola, I need to fix the bottom-most point at the origin, so my equation changes as $$ x^2+\left(\frac{y-b}{b}\right)^2=1.$$ Here again I let $b$ get really really big, but now I have no idea how to explain that what I get is not $$ x^2+1=1.$$ Because, when you draw the picture, it is clear that as $b$ gets bigger and bigger, you get closer to a parabola. But how can I show this algebraically, without going into a whole thing on limits, etc.? [This is a precalculus class.]

To summarize, my question is

How do I show the equation $x^2+\left(\dfrac{y-b}{b}\right)^2=1$ eventually becomes the equation of a parabola as $b\rightarrow\infty$, intuitively? (No limits, no formal arguments please.)

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You don't, since it doesn't. Whatever playing around you do with the $y$ part, we must have $|x|\le 1$. – André Nicolas Apr 16 '13 at 1:54
Well I think you do get $x^2+1=1$ because $[(y-b)/b]^2$ can be rewritten as $[(y/b)-1]^2 and the limit as b approaches infinity is 1. – Ovi Apr 16 '13 at 20:37

You'll need to check details, bu the standard picture is a circular cone, a fixed point partway along its axis, and the plane you are calling the xy plane rotating while always going through that fixed plane. In the beginning, with the plane orthogonal to the axis, you get a circle. As the plane gets closer and closer to parallel to a fixed generating line of the cone, the intersection of the cone with the plane gets closer to, and eventually becomes, a parabola.

Or, a conic section is a circle drawn on a unit sphere that is allowed to move around in space, and we use central projection onto the xy plane. When a point of the circle reaches horizontal, so that the radius of the sphere is parallel to the xy plane, the result is a parabola

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I did the first analysis myself way back when I was in Precalculus. (It's one of the things that got me interested in math research.) You can write some nice equations in terms of the "cone angle" and "cutting plane angle" by keeping the origin of the cutting plane (hence, the origin relative to the conic's equation) at the point where the plane meets the cone axis. Compare these to equations you get by keeping a focus at the origin. Also, compare the equation of the directrix with the equation of the line where the cutting plane meets the plane perpendicular to the cone axis at the apex. – Blue Apr 16 '13 at 2:21
@Blue, my god, you are Apollonius of Perga. How have you survived all these years? – Will Jagy Apr 16 '13 at 2:26
@Blue, meanwhile, I've never worked out how things such as the directrix work out with respect to either the cone or moving sphere model. So that's why I said details needed work. However, it is a nice non-formal picture. – Will Jagy Apr 16 '13 at 2:35
shhhhhhh!!! ... I owe Archimedes money. I don't want him to know I'm here. – Blue Apr 16 '13 at 2:40
@Blue, our little secret. – Will Jagy Apr 16 '13 at 2:57

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