The parabola $x^2=12y$ rolls without slipping around the parabola $x^2=-12y$

Question

The parabola $x^2=12y$ rolls without slipping around the parabola $x^2=-12y$ then find the locus of focus of rolling parabola and also find the locus of vertex of rolling parabola

Answer 1

\begin{align} \text{Static parabola: }f(x)&=-\tfrac1{12}x^2 ,\\ f'(x)&=-\tfrac1{6}x . \end{align}

Tangent line at $x=x_t$: \begin{align} f_t(x,x_t)&= f'(x_t)\,(x-x_t)+f(x_t) \\ &= -\tfrac1{6}x_t\,(x-x_t)-\tfrac1{12}x_t^2 \\ &= -\tfrac16\,x_t\,x+\tfrac1{12}\,x_t^2 . \end{align}

\begin{align} |OD|&=\tfrac12x_t \quad\text{(why?)} ,\\ \tan\theta&=\tfrac16\,x_t ,\\ \sin\theta&= \frac{\tan\theta}{\sqrt{1+\tan^2\theta}} =\frac{x_t}{\sqrt{x_t^2+36}} ,\\ \cos\theta&= \frac{6}{\sqrt{x_t^2+36}} ,\\ |OO'|&=2|OD|\sin\theta =\frac{x_t^2}{\sqrt{x_t^2+36}} ,\\ O'_x&=|OO'|\,\cos(90^\circ-\theta)=|OB|\,\sin\theta =\frac{x_t^3}{x_t^2+36} ,\\ O'_y&=|OO'|\,\sin(90^\circ-\theta)=|OB|\,\cos\theta = \frac{6x_t^2}{x_t^2+36} .\\ \end{align}

Hence, the locus of vertex of rolling parabola is defined parametrically as

\begin{align} O'(x_t)&= \left(\frac{t^3}{t^2+36}, \frac{6t^2}{t^2+36}\right) ,\quad t\in\mathbb{R} . \end{align}

This can be used as a base to derive the locus of the rolling focus.

Answer 2

Let $A=(x, x^2/12)$ be a point on the initial position of the rolling parabola: it will eventually touch the static parabola at $A'=(x, -x^2/12)$. The tangent at $A'$ is the bisector of $\angle CA'A$, where $C=(-3,0)$ is the focus of the static parabola and the rolling parabola (dashed blue in the diagram) is the reflection of the static parabola about the tangent.

The focus $C'$ of the rolling parabola is then the reflection of $C$ about the tangent: it thus belongs to line $AA'$ and $A'C'=A'C$. That means that $C'$ belongs to the directrix of the static parabola, i.e. to line $y=3$, which is then the desired locus.

To find the locus of $V'$, the vertex of the rolling parabola, notice that it is the midpoint of $C'F$, perpendicular to the directrix $CF$ of the rolling parabola. It follows that $C'V'$ is parallel to $CA'$ and $C'V'=CV=3$. We can then compute: $$ V'=C'+3{\vec{A'C}\over A'C}=(x,3) +{3\over 3+x^2/12}(-x,-3+x^2/12)={1\over 36+x^2}(x^3, 6x^2). $$ This is the parametric equation of a Cissoid of Diocles generated by the circle of center $(0,3)$ and radius $3$.