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I have a simple parabola in the form $y = a + bx^2$. I would like to find the formula for a curve which is parallel to this curve by distance $c$. By parallel I mean that there is an equal distance along a line perpendicular to the tangent to my curve at all points.

I've established that the curve isn't in the form $y = a + c + dx^2$, whilst I can make this satisfy for $x=0$ and $x$ equal to one other number is isn't valid across the range.

Any help much appreciated.


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$a+bx$ is a line and not a parabola. – J. M. Sep 1 '11 at 14:16
$y=a+bx$ is a straight line, not a parabola. $y=c+bx$ would be parallel to it. A parabola might be of the form $y=a+ bx +cx^2$. – Henry Sep 1 '11 at 14:17
Sorry, my mistake. I should be y = a + bx^2. I'll edit the question. – Rob Sep 1 '11 at 14:31
up vote 8 down vote accepted

I'll use the parametrization


where $a$ is the focal length (the distance from vertex to focus).

Using the formula for a parallel curve of $(f(t)\quad g(t))^T$ at a distance $c$:


we find the parametric equations


The corresponding Cartesian equation is rather complicated:

$$\begin{align*}x^2 \left(-8 a^3 y+x^2 \left(a^2-10 a y-3 c^2+y^2\right)-20 a^2 c^2+32 a^2 y^2+2 a c^2 y-8 a y^3+3 c^4-2 c^2 y^2+x^4\right)&=\\(c-y) (c+y) \left(4 a(a-y)+c^2\right)^2\end{align*}$$

so you're better off sticking to a parametrization.

Here's a plot of a bunch of parabola parallels:

parabola parallels

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For $y=u+vx^2$, it suffices to consider the case $u=0$ since nonzero $u$ corresponds to a vertical translation. We then have the equivalence $v=\frac1{4a}$. – J. M. Sep 1 '11 at 14:47
Many thanks - that's really helpful. Could you advise how I can utilise the formula in (for example) Excel? – Rob Sep 1 '11 at 14:57
@Rob: If you pick out a value of (in my notation) $a$ and $c$, you can generate a bunch of $(x,y)$ values corresponding to a given $t$, which you can scatter-plot or something... – J. M. Sep 1 '11 at 15:01
Thanks. That's great. – Rob Sep 1 '11 at 21:51

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