If I know the dimensions of a right circular cone (base radius, height and slant height; and place an elliptical cross section inside the cone; how do I get the equations for the original ellipse and what does it look like in the cone's UNFOLDED plane diagram?
Problem: A right circular cone stands on a horizontal base of radius $r$. Its vertex $V$ is a distance $k$ from each point on the perimeter of the base. The plane section of a cone is an ellipse with lowest point $L$ and highest point $H$. On the curved surface of the cone, to one side of plane $VLH$, two routes $R_1$ and $R_2$, from $L$ to $H$ are marked; $R_1$ follows the semi-perimeter of the ellipse, while $R_2$ is the route of shortest length. Find the condition that $R_1$ and $R_2$ intersect between $L$ and $H$. (No answer online. From 1975 British Math Olympiad; problem 5).
I really want to go from an ellipse as the plane section mentioned drawn on a right circular cone to unravelling it into its plane net - to "see" if the $R_1$ and $R_2$ actually intersect. Will this occur for a "narrow / tall" cone or a "wide ' short" cone?
As it was to be expected from a Math Olympiad problem, there is a fairly simple solution to this question.
The semi-ellipse $R_1$ (blue in the unfolded cone represented below) forms a right angle with generatrices $VL$ and $VH$, at points $L$ and $H$ (this is obvious if you consider the 3D setting). Consider now the angles formed by the geodesic $R_2$ with the same generatrices at $L$ and $H$: they can be better visualised on the unfolded surface of the cone, where the geodesic is a straight line (red in figure below). We always have $\angle VLH<90°$, because $VL>VH$. On the contrary, $\angle VHL$ can be acute but also obtuse.
But if $\angle VHL>90°$, then the geodesic joins the ellipse at $H$ "from below", implying an intersection between the curves. The limiting case is $\angle VHL=90°$, which is attained for $VH=VL\cos\beta$ (see figure).
But $\beta k=\pi r$, because the length of the circular arc in the unfolded figure is the same as the perimeter of the base of the cone. Hence the curves intersect if: $$ {VH\over VL}<\cos{\pi r\over k}. $$
EDIT.
To help understanding, here are two figures representing the other two cases: $\angle VHL<90°$ (left) and $\angle VHL>90°$ (right), with the curves intersecting in the latter. All figures were made with GeoGebra Classic 5.