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.)

 A: 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
A: The matter is that you shall keep one of the vertices in $y$ ( for $x=0$) finite and constant.
So starting from the circle
$$
{{x^{\,2} } \over {r^{\,2} }} + {{y^{\,2} } \over {r^{\,2} }} = 1
$$
rewrite it as
$$
{{x^{\,2} } \over {r^{\,2} }} = 1 - {{y^{\,2} } \over {r^{\,2} }} = \left( {1 + {y \over r}} \right)\left( {1 - {y \over r}} \right)
$$
We want to keep the solution $(0,-r)$ for instance, so
$$
{{x^{\,2} } \over {r^{\,2} }} = \left( {1 + {y \over r}} \right)\left( {1 - {y \over {br}}} \right)
$$
for $b \in [1, \infty)$ will give you circle, ellipse, parabola.
Clearly , if you stretch instead both roots, you get a centered ellipse, which in the limit becomes two parallel lines.
A: In the equation you have to fix the angle under you see the "width" of the ellipse (its axis parallel to $x$) from the origin. Or else as $b$ approaches infinity the apparent "width" of your ellipse from the origin, approaches zero. So we have to scale $x$ in proportion how much we move the center of the ellipse. This is one way to do this (but you can write any constant above $b$, even $a$):
$$
{\frac 1 b}{\Bigl(\frac x a\Bigr)^2}+\Bigl(\frac{y-b}b\Bigr)^2=1
$$
Now if you substitute:
$$
\frac 1 b=c
$$
Then your equation becomes:
$$
c{\Bigl(\frac x a\Bigr)^2}+(cy-1)^2=1
$$
This can be simplified to:
$$
{\Bigl(\frac x a\Bigr)^2}+cy^2-2y=0
$$
As $b$ approaches infinity, $c$ approaches zero. And now it is apparent that if $c$ approaches zero, the equation approaches some parabola's equation. It is nice to notice that when $c$ is negative, the equation becomes a hyperbola's equation. You can think of $c$ as something related to the angle of the plane @Will explained in his answer.
