## Parabola Extension

- 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...?

3.14159265358979323846264338327950288419716939937510

58209749445923078164062862089986280348253421170679...?

### 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 )

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?

...

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...?

3.14159265358979323846264338327950288419716939937510

58209749445923078164062862089986280348253421170679...?

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

### Re: Parabola Extension

Thestijn263 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 )

**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 gypaètes sont là.

- 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...?

3.14159265358979323846264338327950288419716939937510

58209749445923078164062862089986280348253421170679...?

### Re: Parabola Extension

Hehe, I'll take that as a complimentelendiastarman wrote:........wow.........

...

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

w00t...

Seriously, how's it coming?

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.

hm, I thought that was theDaniel wrote:Theeasiestcase 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.

**trivial**case, but it's an easy case no matter what you call it

### Re: Parabola Extension

Why would you want that?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.

Isn't a prarametric equation like x=f(t), y=g(t) enough to describe the locus?

### Re: Parabola Extension

Yes it is. But I have the following equations: f(t,x,y) = 0, g(t,x,y) = 0Why would you want that?

Isn't a prarametric equation like x=f(t), y=g(t) enough to describe the locus?

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

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

### 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))

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))

### 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..

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..

### 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.

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.