Sorry about the variables. I must have changed how I wanted to represent my solution, but didn't finish the job. But it doesn't matter, cause that's not what you wanted!

I misunderstood what you wanted, but now I think I get it. Unfortunately, I'm running out and won't have time for a few days. If you still haven't gotten what you need by next week, I'll give it another go.

I actually didn't post anything new recently. I don't know why it says I did. But anyway, it's easy to add a line. Starting with Gianluca's version, just add a graphics element with an InfiniteLine:

hyperbola = 
  ContourPlot[{-10 x - x^2 + (2 y)/Sqrt[3] + 2 Sqrt[3] x y + y^2 == 0,
     y == 2/Sqrt[3] - x/Sqrt[3], y == Sqrt[3] x + 8/Sqrt[3]}, {x, -9, 
    7}, {y, -6, 10}, ContourStyle -> Directive[Black, Dotted], 
   ImageSize -> 20.1 16.1];
sliderHyperbola[p_] := 
  ContourPlot[-10 x - x^2 + (2 y)/Sqrt[3] + 2 Sqrt[3] x y + y^2 == 
    0, {x, -14, 9}, {y, -6, 10}, ContourStyle -> Red, 
   RegionFunction -> Function[{x, y}, y > x Sqrt[3] - p]];
Manipulate[
 Show[hyperbola, sliderHyperbola[p], 
  Graphics[{Blue, InfiniteLine[{0, -p}, {1, Sqrt[3]}]}]], {p, -14, 8}]

Just style it however you want.

ImageSize specifies the size in printer points. Your 16 by 16 is very very small. The picture becomes visible because you wrote 20.1 16.1, which is interpreted as a multiplication.

I have no idea why you cannot reproduce the Manipulate.

Your hyperbola does not depend on p, so that it is better to pre-calculate it before feeding it to Manipulate. I have rearranged things a little in a way that seems a little cleaner to me:

hyperbola = 
  ContourPlot[{-10 x - x^2 + (2 y)/Sqrt[3] + 2 Sqrt[3] x y + y^2 == 0,
     y == 2/Sqrt[3] - x/Sqrt[3], y == Sqrt[3] x + 8/Sqrt[3]},
   {x, -9, 7}, {y, -6, 10},
   ContourStyle -> Directive[Black, Dotted], ImageSize -> 20.1 16.1];
sliderHyperbola[p_] := 
  ContourPlot[-10 x - x^2 + (2 y)/Sqrt[3] + 2 Sqrt[3] x y + y^2 == 
    0, {x, -14, 9}, {y, -6, 10}, ContourStyle -> Red, 
   RegionFunction -> Function[{x, y}, y > x Sqrt[3] - p]];
Manipulate[
 Show[hyperbola, sliderHyperbola[p]],
 {p, -14, 8}]

I hope it doesn't confuse you.

I'm using 13.2 also. I imagine that you, as an advanced programmer, have been able to size up my situation so well that you can just edit one version and not have any conflicts

I have multiple versions of the Manipulates you and others have posted for me, and I think in storage I somehow make the same words have different definitions. I think that's why I get failures. That's why I use Module, correctly or incorrectly.

If there were ever such a thing as a Mathematica "high," I'm in one today, thanks to you, Eric. It was just before Christmas when I had my doubts that it was even possible to get a line to move parallel to itself.

As for your last program with the infinite line, I kept getting an error on execution, even after quitting the kernel. This was the same error I was getting yesterday. I did not have the success reported by Gianluca Gorni.

Eventually it occurred to me that the defined variables were not enclosed in Module. Therefore, I took your new code for the infinite line, and put it in the program that had the Module. I'm writing a paper and I wanted to use this Manipulate to illustrate it. Without your help I never would have developed the Manipulate. About all I did was furnish the idea.

Thank you for bringing into existence what I could only imagine. Here is how I put the variables inside Module:

Manipulate[
 Module[{hyperbola, sliderHyperbola, sliderLine}, 
  hyperbola = 
   ContourPlot[{-10 x - x^2 + (2 y)/Sqrt[3] + 2 Sqrt[3] x y + y^2 == 
      0, y == 2/Sqrt[3] - x/Sqrt[3], 
     y == Sqrt[3] x + 8/Sqrt[3]}, {x, -14, 9}, {y, -6, 10}, 
    ContourStyle -> {{Directive[Black, Dotted]}}]; 
  sliderHyperbola = 
   ContourPlot[-10 x - x^2 + (2 y)/Sqrt[3] + 2 Sqrt[3] x y + y^2 == 
     0, {x, -14, 9}, {y, -6, 10}, ContourStyle -> Directive[Red], 
    RegionFunction -> MyRegionFunction[p]];
  sliderLine = Graphics[{Blue, InfiniteLine[{0, -p}, {1, Sqrt[3]}]}];
  Show[hyperbola, sliderHyperbola, sliderLine, Frame -> True, 
   ImageSize -> 400.1, AspectRatio -> Automatic, 
   PlotRange -> {{-9.5, 6.5}, {-6, 10}}]], {p, -28, 19, 0.025}, 
 ContinuousAction -> True]

Yes, when I quit the kernel the Manipulate worked. As for the Image function, I wasn't thinking clearly when I tried to answer the question. I had Image calling for a 320x320 square, but when I wanted to center the hyperbola with the right amount showing, I used PlotRange.

Thank you for your interest and observations.

Appreciate your quick response. Unfortunately, I'm only getting pink output. It seems the two plots won't combine. If you paste back from the browser do you get a working output?

Above, in my response to Eric Rimbey, I referred to you as "Upgrading Home"! How bad is that? I got your first name and last name wrong! My humble apologies.

I was asking Upgrading Home to rewrite his definition using pure functions:

MyRegionFunction[p_] := #2 > #1 Sqrt[3] - p &

in the clumsy way I would try to write without pure functions a definition that would do the same thing.

Nevertheless, it is you that got me moving ahead immediately in answering my first question a couple of weeks ago. All of your coding was of the greatest enlightenment for me because it was tailored to my focus. Thanks for getting me to where I am now so quickly.

I watched the pro football players so excited with their victories, sometimes living in dreams beyond their expectations. Believe me, after struggling without success for so long, when I watched the output of your program (and Upgrading Name's) unfold, I felt pretty good myself.

I doubt that anyone else will offer a solution for me. I feel unbelievably lucky that you found something in my program to devote your time and attention to it. Even if I never get any more code from you, I will remember how much I've learned from your efforts.

Anything you send my way will be more beneficial than anything I can think of because it will be customized to me - even if it has mistakes!