Parabola Extension

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elendiastarman
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Parabola Extension

Post by elendiastarman » Fri Aug 14, 2009 9:01 pm

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)?
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stijn263
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Re: Parabola Extension

Post by stijn263 » Wed Aug 19, 2009 8:49 am

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 )

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elendiastarman
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Re: Parabola Extension

Post by elendiastarman » Wed Aug 19, 2009 8:27 pm

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

Seriously, how's it coming?
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daniel.is.fischer
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Re: Parabola Extension

Post by daniel.is.fischer » Wed Aug 19, 2009 9:14 pm

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 gypaètes sont là.

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elendiastarman
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Re: Parabola Extension

Post by elendiastarman » Thu Aug 20, 2009 12:19 am

Well, at least we can run an imprecise simulation and see that the shape is something like this:
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stijn263
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Re: Parabola Extension

Post by stijn263 » Fri Aug 21, 2009 6:32 pm

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

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

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

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hk
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Re: Parabola Extension

Post by hk » Fri Aug 21, 2009 6:46 pm

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?
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stijn263
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Re: Parabola Extension

Post by stijn263 » Sat Aug 22, 2009 12:10 pm

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

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hk
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Re: Parabola Extension

Post by hk » Sat Aug 22, 2009 9:03 pm

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))
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stijn263
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Re: Parabola Extension

Post by stijn263 » Sun Aug 23, 2009 1:22 pm

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..

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hk
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Re: Parabola Extension

Post by hk » Sun Aug 23, 2009 2:08 pm

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.
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