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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.

Rob

Stahl
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Rob
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1 Answers1

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I'll use the parametrization

$$\begin{align*}x&=2at\\y&=at^2\end{align*}$$

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 $c$:

$$\begin{pmatrix}f(t)\\g(t)\end{pmatrix}+\frac{c}{\sqrt{f^\prime(t)^2+g^\prime(t)^2}}\begin{pmatrix}g^\prime(t)\\-f^\prime(t)\end{pmatrix}$$

we find the parametric equations

$$\begin{align*}x&=2at+\frac{ct}{\sqrt{1+t^2}}\\y&=at^2-\frac{c}{\sqrt{1+t^2}}\end{align*}$$

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

J. M. ain't a mathematician
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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. ain't a mathematician Sep 01 '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 01 '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. ain't a mathematician Sep 01 '11 at 15:01