# Turning an ellipse into a parabola

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