Given a known vertical, negative parabola ($y=ax^2+bx+c$ with $a$,$b$ & $c$ known parameters): how far down from the vertex of the parabola (the top) is the vertical distance, so when you "cut if off" with a horizontal line, this parabola segment is still a 3D-rotated circle segment? I refer to the following image:
In step $1$ a frontal view of a circle can be seen in a 3-dimensional space on $X$, $Y$ & $Z$ axes: Green is the $X$-axis, Blue the $Y$-, and red the $Z$-axis: as this is a 0 degree frontal view, $Z$ & $X$ are on the same horizontal line.
In step $2$ we now rotate this circle to the right (or left: does not matter) with a certain $\theta$ degrees. So it's still the same circle, and the same point of view, but the circle is rotated around its vertical ($Y$) axis.
In step $3$ a parabola can be seen for which the top part (above the red horizontal line $AB$) strongly resembles the top part (above the $AB$-line) of the rotated circle in the previous step. Please bear in mind that this image is an approximation.
The horizontal $AB$-line can only go so low. This is the distance $X$ in step $1$. This distance is also indicated by the rectangular boxes in point $A$ and point $B$ in step $2$: this is the point where the rotated circle on the left and on the right side stop moving away from the axis of rotation, and again start coming closer to it (sounds a bit like a derivative I guess). Could be related to conic sections also. This is the opposite of / related to this question.