1

Just a little background, I'm a maths teacher in highschool, and this is not a problem I deal with on the daily. I just came across it trying to create graphics for class and I've been stuck.

So, my goal is to find a function (not just the graph, a formula) that at has the distance of 1 to any given point of the quadratic function. Of course, there are two of these functions, one "inside" and one "outside". Important is that the distance between the two graphs is measured along the normals of the quadratic function.

I first started calculating some points and then interpolate a polynomial, but soon had to see that this doesn't work.

Now, I worked out the universal formula for any normal line through any point $P(x_0|f(x_0))$ of the quadratic function, which is

$y_n=- \frac{1}{2 x_0}x+x_0^2+ \frac{1}{2}$ with $x_0 \neq 0$

Now, we could intersect those normal lines with circles with radius 1. A circle with radius 1 and center $P(x_0|f(x_0))$ has the formula

$1=(x-x_0)^2+(y+x_0^2)^2$

These two formulas will give us a nonlinear system of equations. I substitute y in the second formula and now I basically have a quadratic equation that I should be able to solve with the p-q-formula. After a little bit of clean up, I had:

$0=x^2- \frac{2x_0}{4x_0^2+1}x+ \frac{16x_0^6+12x_0^4-3x_0^2}{4x_0^2+1}$

Plugging that into the p-q-formula and cleaning up a little, this is my radicand: $\frac{4x_0^2-64x_0^8-64x_0^6}{(4x_0^2+1)^2}$

I need to take the root of that, but I can't seem to find an expression of the numerator as a square.

Could someone give me some insight?

Or: What do you think of my approach to find the solution to the problem? Is there something more efficient?

I do have a pdf where I wrote down everything much more detailed, if anyone's interested...

Ellon Feyneth
  • 21
  • What is your $f$? Is it $f(x)=x^2$? Also, what is the p-q formula? Posting your PDF would be helpful. – Benjamin Wang Oct 22 '23 at 21:01
  • Yes indeed. $f(x)=x^2$. Just the normal quadratic function.

    And here is the link to my pdf: https://drive.google.com/file/d/11QQKRzjsSz6Ty4BU2u4WPiiy-cWX4GzZ/view?usp=sharing

    Oh and the p-q-formula is a formula that solves quadratic equations, if the quadratic equation has the form $0=x^2+px+q$ with $x_{1/2}=-\frac{p}{2}+-sqrt((\frac{p}{2})^2-q)$

     – Ellon Feyneth Oct 22 '23 at 21:06
  • Oooh, great! Yeah, I didn't think of the word "parallel". Thanks for suggesting that. I'll work through that problem with parallel curves in mind. I'll have to read up on parametrically defined curves/functions. Thanks! – Ellon Feyneth Oct 22 '23 at 21:41

1 Answers1

1

Okay, thanks to Amateur_Algerist, here's the answer to the problem:

Define parameter t that runs through the real numbers.

Then define the original function parametrically with $P( t | t^2) $ ($x(t)=t$ and $y(t)=t^2$)

All points $P_g$ can thus be expressed (through the formula of parallel curves) with $P_g(t+ \frac{2t}{\sqrt{1+4t^2}} | t^2 - \frac{1}{\sqrt{1+4t^2}})$ or $P_g(t- \frac{2t}{\sqrt{1+4t^2}} | t^2 + \frac{1}{\sqrt{1+4t^2}})$

I'm gonna try and work on deparameterizing that form, so when/if I have a solution to that, I'll edit the answer.

Thanks to all for your help.

Ellon Feyneth
  • 21