TITLE: The parabola $x^2=12y$ rolls without slipping around the parabola $x^2=-12y$
QUESTION [7 upvotes]: 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

REPLY [2 votes]: 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 $3$.