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What would be the name of the shape that is the set of all points such that they are equidistant from the point $(0,1)$ and to the parabola $y=x^2$.

Figure

Here is a desmos graph that generates the points. It is basically using the same process that creates a parabola with a focus and a directrix but this directrix is a parabola rather than a line

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    $\begingroup$ I don't know a name of the shape. Looks similar to the curve $$x^2 = \left(y-\dfrac{1}{2}\right)\left(y-\dfrac{3}{2}\right)^2$$ Link to Wolfram Alpha:wolframalpha.com/input/… $\endgroup$ – David Peterson Oct 9 '14 at 3:08
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    $\begingroup$ I chugged through some annoying algebra and wound up with $$ x = \frac{2t^3-t^5}{t^2+1}, \quad y = \frac{3t^4+t^2+1}{2t^2+2} $$ Not sure if I made a mistake, but it looks close... and pretty hard to see any nice form/name for. wolframalpha.com/input/… $\endgroup$ – Platehead Oct 9 '14 at 5:25
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    $\begingroup$ The implicit form of the curve is $$8x^2y^3 + 32y^5 - 27x^4 - 156x^2y^2 - 208y^4 + 150x^2y + 464y^3 - 37x^2 - 424y^2 + 170y - 25=0$$ i.e. an algebraic curve of degree 5. Therefore the equation of degree 3 David wrote is really just similar, not the same. @Platehead: I can confirm your parametric form. $\endgroup$ – MvG Oct 9 '14 at 7:57
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Given a point $P$, if $Q$ is the nearest point on the parabola to $Q$, then $P$ lies on the normal through $Q$. Unfortunately, the converse is not the case: depending on its position, $P$ may lie on up to four different normals, and usually only one of these leads to the closest point.

For instance, in your diagram, the point on the red curve is actually closer to the right-hand arm of the parabola than the left-hand arm, so its distance to the parabola is less than is shown in the diagram.

Having said that, we can find the equation of the red curve as follows:

Let $Q$ be a point $(t,t^2)$ on the parabola, and let $P$ be a point on the normal through $Q$, equidistant from $Q$ and $(0,1)$. Then $P$ lies on the perpendicular bisector of $Q$ and $(0,1)$. So all we have to do is find where the normal and the perpendicular bisector intersect.

The normal goes through $Q$, and has a slope of $-1/2t$, so its parametric form is $(t+2t\lambda,t^2-\lambda)$ ($\lambda \in \mathbb R$).

The perpendicular bisector goes through $(t/2,(t^2+1)/2)$, and has a slope of $t/(1-t^2)$, so its points are $(t/2+\mu(1-t^2),(t^2+1)/2+\mu t)$ ($\mu \in \mathbb R$).

Equate the $x-$ and $y-$coefficients to find the point of intersection:

$$t+2t\lambda = t/2+\mu(1-t^2)$$ $$t^2-\lambda = (t^2+1)/2+\mu t$$

From the second equation, $\lambda = (t^2-1)/2-\mu t$. Plugging this into the first equation gives $$t^3-2\mu t^2 = t/2 + \mu(1-t^2)$$ which rearranges to $$\mu = \frac{t(2t^2-1)}{2(t^2+1)}$$

Substituting this value for $\mu$ in the point on the perpendicular bisector gives a parametric form for the red curve:

$$\left(\frac{t^3(2-t^2)}{2(t^2+1)},\frac{3t^4+t^2+1}{2(t^2+1)}\right)$$

This passes the Wolfram Alpha eyeball test, which means it's possible that I haven't made any errors. But even if that's the case, the curve probably has a simpler form. (On the other hand, it looks suspiciously as if the curve is not differentiable at $(0,\frac12)$, so perhaps a complicated equation is inevitable.)

Edited to add: Oh $-$ I see now that you wanted the name of the curve, not its equation. That I don't know.

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  • $\begingroup$ The equation is as good as the name. All I wanted was information about it and this answer has what I needed. $\endgroup$ – Jacob Jan 12 '15 at 3:25
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I am pretty sure this is called a "right strophoid". Seems pretty similar.

http://mathworld.wolfram.com/RightStrophoid.html

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  • $\begingroup$ This is pretty cool. When the directrix is a line (degree 1), you get a parabola (degree 2). When the "pseudo-directrix" is a parabola (degree 2), you get a strophoid (degree 3). $\endgroup$ – dardeshna Dec 22 '14 at 22:34

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