Are Parabolas similar intuitively?

Question

All parabolas are similar, but are they all similar in that it is just a question of 'zooming in and out' intuitively speaking? It seems that there should therefore be on all parabolas a curve from the origin to a 'scaled up or down' version of the point (1,1), in order for them to 'fit on top of one another' (the way different scaled circles or squares would). Re Matthews request. When we say circles (and squares) are similar none of my students have any intuitive difficulty with this. With parabola it seems to them not to be the case. Many feel that scaling 'distorts' the parabolas and cannot intuitively grasp how a parabola and, for example, half a hyperbola differ. The textbooks don't help, and without an (better) intuitive 'handle' the algebraic approach will be only followed by the brighter guys. I'm looking for a better place to start, which will help me take more students further (on the road to a good grasp of conics.)

Answer 1

If you are only talking about $y$ as a function of $x$ for example in a 2 dimensional cartesian grid, then scaling is not enough. Let's say $f(x)=x^2$ is the "basic" parabola. Then to get any parabola you can take our basic parabola and then

1. Scale it - which takes care of stretching/shrinking the parabola
2. Move it horizontally
3. Move it vertically
4. And multiplication by $-1$ which determines if the parabola opens up or down. You can combine this with number 1, if you allow scaling constants to be negative.

Now, if you were asking about ANY parabola, which includes for example $y=x^2$, $x=y^2$, then in addition to the first four, you also need

5.Rotation - where you can rotate the parabola clockwise/counterclockwise by any angle around the origin let's say.

With these five operations, now you can start with $y=x^2$ and obtain any other parabola.

So for example,

1. Start with $y=x^2$, rotate it by 90 degrees ccw and you get $x=y^2$.

2. Start with $y=x^2$, multiply by -3, move it up 3, move it 4 to the right and you get $y=-3(x-4)^2+3=-3x^2+24x-45$.

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Addendum: Per Mathwood's comment

If you have a parabola $y=ax^2+bx+c$ and you want it to go through the point $(k,k)$ then just multiply it by the scalar

$$\frac{k}{ak^2+bk+c}$$

so your new parabola will be

$$y=\frac{k}{ak^2+bk+c}(ax^2+bx+c)$$ and this works for any parabola as long as of course $ak^2+bk+c\neq0$. Meaning that even if the original parabola doesn't go through (1,1), this new parabola will go through $(k,k)$.