A parabola is defined as the set of points equidistant from a directrix line and a focus point. But what if we allow the directrix to be an arbitrary curve, including possibly another parabola? What happens?
My attempts so far...
Let $(f, \phi)$ represent the focus point F. Let $(q, Q)$ represent points on the directrix curve K. Let $(x,y)$ represent a point P on the curve that satisfies the condition : $$distance (FP) = distance (KP)$$
By applying the condition, we get the following relation between F, K and P:
$$y = \frac{(x-f)^2 - (x-q)^2}{2(\phi-Q)} + \frac{\phi+Q}{2} \hspace{10mm}E1 $$
This comes from applying the distance formula to the above condition, then expanding and simplifying. Let’s test this expression on the directrix curve $K=(x,-1)$ and the focus $F=(0,1)$. Plugging this into the above expression, we get $$y = \frac{(x-0)^2 - (x-x)^2}{2(1-(-1)} + \frac{1+(-1)}{2} = \frac{x^2}{4} $$ Which works. But for some reason, I can’t make sense of E1. Is it general enough? How can I use it for parabolas tilted at some angle e.g. if the directrix is the line defined by $K(q,Q) = (q,2q)$ and the focus is at $F(f,\phi) = (0,1)$ then intuitively the parabola should be slanted. But how can I describe such a parabola with E1, if E1 is correct?
Further motivation - what if the directrix curve is $K(q,Q) = (q,cos(q))$ and the focus is at $(0,1)$? What curve described by $(x,y)$ satisfies E1?
Not an answer: just to show the locus proposed by the OP in a comment (easily made with GeoGebra):
The same locus when the "directrix" is parabola $y=x^2$, and the focus the focus of the parabola.