## Parabola Extension

Shape and space, angle and circle properties, ...
elendiastarman
Posts: 410
Joined: Sat Dec 22, 2007 8:15 pm

### Parabola Extension

The locus of all points equidistant between a point and a line is a parabola. What kind of curve is the locus of all points equidistant between a point and a parabola (referred to earlier)?
Want some
3.14159265358979323846264338327950288419716939937510
58209749445923078164062862089986280348253421170679...?

stijn263
Posts: 1505
Joined: Sat Sep 15, 2007 10:57 pm
Location: Netherlands

### Re: Parabola Extension

Thanks for a fun problem! I haven't solved it yet, but I'm getting somewhere I think

I'm investigating the easiest case ( a point on x = 0 and y = y0 and the parabola y = x^2 )

elendiastarman
Posts: 410
Joined: Sat Dec 22, 2007 8:15 pm

### Re: Parabola Extension

........wow.........
...
I CREATED A PROBLEM THAT STIJN263 HAS BEEN WORKING ON FOR DAYS AND THINKS ITS FUN?
w00t...

Seriously, how's it coming?
Want some
3.14159265358979323846264338327950288419716939937510
58209749445923078164062862089986280348253421170679...?

daniel.is.fischer
Posts: 2400
Joined: Sun Sep 02, 2007 10:15 pm
Location: Bremen, Germany

### Re: Parabola Extension

stijn263 wrote:Thanks for a fun problem! I haven't solved it yet, but I'm getting somewhere I think

I'm investigating the easiest case ( a point on x = 0 and y = y0 and the parabola y = x^2 )
The easiest case is when the point is on the parabola. Then we have a ray orthogonal to the parabola and the only difficulty is calculating how far into the interior it extends.
Il faut respecter la montagne -- c'est pourquoi les gypa&egrave;tes sont l&agrave;.

elendiastarman
Posts: 410
Joined: Sat Dec 22, 2007 8:15 pm

### Re: Parabola Extension

Well, at least we can run an imprecise simulation and see that the shape is something like this:
Want some
3.14159265358979323846264338327950288419716939937510
58209749445923078164062862089986280348253421170679...?

stijn263
Posts: 1505
Joined: Sat Sep 15, 2007 10:57 pm
Location: Netherlands

### Re: Parabola Extension

elendiastarman wrote:........wow.........
...
I CREATED A PROBLEM THAT STIJN263 HAS BEEN WORKING ON FOR DAYS AND THINKS ITS FUN?
w00t...

Seriously, how's it coming?
Hehe, I'll take that as a compliment

Anyways, I'm not working on it all the time, but it's not an easy problem. I'm trying to solve it with pen and paper, but I have 2 polynomial equations with 3 unknowns, and I need to reduce them to 1 equation with 2 unknowns (in x and y). I'm having some difficulties doing that Perhaps I'll try to solve it in Maple later.
Daniel wrote:The easiest case is when the point is on the parabola. Then we have a ray orthogonal to the parabola and the only difficulty is calculating how far into the interior it extends.
hm, I thought that was the trivial case, but it's an easy case no matter what you call it

hk
Administrator
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Joined: Sun Mar 26, 2006 9:34 am
Location: Haren, Netherlands

### Re: Parabola Extension

stijn263 wrote: Anyways, I'm not working on it all the time, but it's not an easy problem. I'm trying to solve it with pen and paper, but I have 2 polynomial equations with 3 unknowns, and I need to reduce them to 1 equation with 2 unknowns (in x and y). I'm having some difficulties doing that Perhaps I'll try to solve it in Maple later.
Why would you want that?
Isn't a prarametric equation like x=f(t), y=g(t) enough to describe the locus?

stijn263
Posts: 1505
Joined: Sat Sep 15, 2007 10:57 pm
Location: Netherlands

### Re: Parabola Extension

Why would you want that?
Isn't a prarametric equation like x=f(t), y=g(t) enough to describe the locus?
Yes it is. But I have the following equations: f(t,x,y) = 0, g(t,x,y) = 0

f = 2*t^3 + t*(1-2*y) - x
g = t^4 + t*x - y0^2 + 2*y0*y

hk
Administrator
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Joined: Sun Mar 26, 2006 9:34 am
Location: Haren, Netherlands

### Re: Parabola Extension

If (a,b) is the point and (t,t^2) is a point on the parabola then I get as point on the locus:
x=(-t*(t^4 - 2*b*t^2 + a^2 + b*(b - 1)))/(t^2 - 2*a*t + b)
y=(3*t^4 - 4*a*t^3 + t^2 - 2*a*t + a^2 + b^2)/(2*(t^2 - 2*a*t + b))

stijn263
Posts: 1505
Joined: Sat Sep 15, 2007 10:57 pm
Location: Netherlands

### Re: Parabola Extension

lol, I was being stupid. I don't need to solve for t, but for x and y. And my equations are just linear in x and y

2*t * y + x = 2*t^3 + t
2*y0 * y + t * x = y0^2 - t^4

So:
2*(y0-t^2) * y = y0^2 - 3*t^4 - t^2

y = (b^2 - 3*t^4 - t^2) / (2*(b-t^2))

Hm, I think I messed it up a bit somewhere with minus signs..

hk
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Joined: Sun Mar 26, 2006 9:34 am
Location: Haren, Netherlands

### Re: Parabola Extension

The normal to the parabola has equation
y=-1/(2t)(x-t)+t^2
and the ppb of (a,b) and (t,t^2) has equation
y=-(x-t)/(y-t^2)(x-(a+t)/2)+(b+t^2)/2
i found the intersection of these with a CAS program.