I keep hearing that when one rolls a parabola on a straight line, the focus traces a catenary. I kept trying to find a proof on the Internet, but no dice. How does one prove this to be true?
Calculus – Why Focus of Rolling Parabola Traces a Catenary?
calculusgeometryplane-curves
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PERHAPS THIS HELPS
I've been thinking about your problem for a while. I wouldn't say that I could solve it. However, I have an idea that I would like to share with you.
You certainly remember the Dandelin spheres which may have something to do with your conjecture. The following figure depicts two cones (a black one and a grey one). These cones share the vertical axis and a Dandelin sphere. Tangent to this sphere there is a blue plane whose edge can be seen in the figure:
The blue plane and the black cone intersect in a parabola. Also, the blue plane intersects with the other cone in a hyperbola. The parabola's focus and one of the foci of the hyperbola are common.
My conjecture is that this is how to imagine your couple of parabola and hyperbola with a common focus.
Then, I don't know how to proceed.
It might help you to go back to the original Euclidean geometry. For example, one theorem in 'The Elements' is: A straight line is the locus of all points equidistant from two (distinct) given points" ('locus of points' just means 'the shape all of the points fall upon and/or trace out'). If you transpose to a Cartesian coordinate system, and accept the obvious translation of the Pythagorean Theorem into an equation for distance between points as a function of coordinates, then you can translate this theorem into an equation between two formulae defining the distance of each 'locus' point from each of the given points. That equation can be simplified and rearranged into the usual linear expression.
Similarly, for Euclid a parabola is -defined- to be 'the locus of all points equidistant from a given line ('directrix') and a given point ('focus') which is not on that line'. Restrict the case to a horizontal line [greatly simplifying the equation for the distance from a point to that line] and a point above the line, then translate the definition into an equation of two distance formulae, and you get a form of the quadratic equation.
Best Answer
Let's roll. :)
Deriving the parametric equations of a straight line roulette isn't terribly complicated. As I mentioned in this previous answer, rolling is best decomposed as a rotation and a translation. For this case, I'll take the straight line to be the horizontal axis.
Let's start again with a convenient parabola parametrization:
$$\begin{pmatrix}2at\\at^2\end{pmatrix}$$
where $a$ is the focal length (the distance from vertex to focus). The focus of this parabola is at the point $(0,a)$.
We also require the arclength function for this parametrization of the parabola: $s(t)=a(t\sqrt{1+t^2}+\mathrm{arsinh}(t))$.
The trick to rolling a parabola is to consider the transformations necessary for a point on the parabola to touch an appropriate point on the straight line it is rolling on. The parametrization I have chosen is particularly convenient, in that the vertex of the parabola already touches the horizontal axis at the origin.
Going forward, we can translate the parabola to be rolled so that the intended contact point coincides with the origin. We then perform a rotation such that the parabola is now tangent to the horizontal axis, and then horizontally translate by an amount equal to the parabola's arclength. (A similar derivation is done for the cycloid.)
Mathematically, we perform this sequence of transformations on the point $(0,a)$; here is the translation:
$$\begin{pmatrix}0\\a\end{pmatrix}-\begin{pmatrix}2at\\at^2\end{pmatrix}$$
The rotation then needed is given by the tangential angle rotation matrix. I derived the expression for the parabola in my previous answer, so I won't repeat it here. The only difference from the previous answer is that to go forward, we require a clockwise rotation, and thus we must transpose the tangential angle rotation matrix. This now gives us
$$\begin{pmatrix}\frac1{\sqrt{1+t^2}}&\frac{t}{\sqrt{1+t^2}}\\-\frac{t}{\sqrt{1+t^2}}&\frac1{\sqrt{1+t^2}}\end{pmatrix}\cdot\left(\begin{pmatrix}0\\a\end{pmatrix}-\begin{pmatrix}2at\\at^2\end{pmatrix}\right)$$
Finally, we translate horizontally with the arclength expression given earlier:
$$\begin{pmatrix}a(t\sqrt{1+t^2}+\mathrm{arsinh}(t))\\0\end{pmatrix}+\begin{pmatrix}\frac1{\sqrt{1+t^2}}&\frac{t}{\sqrt{1+t^2}}\\-\frac{t}{\sqrt{1+t^2}}&\frac1{\sqrt{1+t^2}}\end{pmatrix}\cdot\left(\begin{pmatrix}0\\a\end{pmatrix}-\begin{pmatrix}2at\\at^2\end{pmatrix}\right)$$
The parametric equations for the roulette surprisingly simplify to
$$\begin{align*}x&=a\operatorname{arsinh}(t)\\y&=a\sqrt{1+t^2}\end{align*}$$
Eliminating the parameter $t$ yields the usual equation for the catenary, $y=a\cosh\frac{x}{a}$.
Here's a picture I previously did:
A similar derivation can be done to show that the directrix of the same rolling parabola envelopes a reflection about the horizontal axis of the catenary being traced by the focus.