Transformation from Circle to Parabola

Question

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Is there a transform that maps points on the circle $$x^2+(y-R)^2=R^2$$ to points on the parabola $$\frac {x^2}{d^2}+\frac y{\frac {d^2}{2(d-R)}}=1$$ as shown by black arrows in the diagram above (and similar for the left side, although not shown in the diagram)?

Note that the parabola touches the circle at $(\pm p,q)$ and these two points map onto themselves.

(See also this question here)

Answer 1

Geometrically:

The circle and the parabola can be seen as two sections of the same cone, by two planes having the line $y=q$ in common. The circle is obtained when the plane is orthogonal to the axis, and the parabola when it becomes parallel to a generatrix. This plane is rotated to match the first.

For intermediate rotations, you get ellipses that remain tangent at the same points.

Answer 2

There are many ways to do this. Here is one way.

For $-2R\le a\le0$, the equations of the circle and the parabola can be written in the form

1. $x^2+(y-R)^2=R^2$
2. $x^2=ay+\left(R-\frac{a}{2}\right)^2$

and the point of intersection will be $(p,q)$ where

$p=\sqrt{R^2-\dfrac{a^2}{4}}$

$q=R-\dfrac{a}{2}$

Define a function $f(x)$ on the interval $[0,p]$ equal to the positive vertical distance between the two curves for $0\le x\le p$:

\begin{equation} f(x)=\frac{1}{a}\left[x^2-\left(R-\frac{a}{2}\right)^2\right]-\left(R+\sqrt{R^2-x^2}\right) \end{equation} and define a function $g(y)$ on the interval $[0,q]$ equal to the positive horizantal distance between the curves for $0\le y\le q$:

\begin{equation} g(y)=\sqrt{ay+\left(R-\frac{a}{2}\right)^2}-\sqrt{R^2-(y-R)^2} \end{equation}

Then for points $(x,y)$ lying on the right half of the circle define the function

\begin{equation} h(x,y)=\begin{cases} (x,y+f(x))\text{ for } 0\le x< p, y> q \\ (x+g(y),y)\text{ for } 0\le y\le q \end{cases} \end{equation}

Addendum: Equation 2 above comes from writing the parabolic equation as $x^2=ay+b$ then finding the value of $b$ which gives only one point of intersection with the circle. It is then easy to find both $p$ and $q$ in terms of $R$ and $a$.