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I draw a circle on a paper and cut a sector of it to make a cone with it.

then I draw an ellipse on the cone. then I netted the cone to have its circular sector again.

the ellipse's perimeter was like what you see in the picture (I have lost it so I draw a similar thing in Microsoft Paint).

how to find ellipse's perimeter, semi-major and semi-minor axes using it?

please don't tell me find the perimeter with a tool. I want a way.enter image description here

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    "I draw an ellipse on the cone": what does that mean ? Was the ellipse planar ? In any case, your picture is insufficient. –  Mar 02 '20 at 12:47
  • the picture says it. I made a cone from a sector. then I draw an ellipse from one side of the sector to the other side. then I opened the cone to have its sector again. the picture is the sector after opening the cone. not the cone itself. –  Mar 02 '20 at 15:23
  • The picture says nothing. Was the ellipse planar ? How did you draw it ?? –  Mar 02 '20 at 16:19
  • I drew the ellipse all around the cone –  Mar 06 '20 at 13:31
  • I'm guessing the ellipse is in the conic section sense (cross section along a plane)? Do you have any other measurements relating to the ellipse? Right now you only have a single point on the cone 3cm down the cone, which might give you some restrictions on the quantities you're trying to find but not much more. – epimorphic Mar 08 '20 at 18:01
  • Hi @epimorphic yes; you got it right. but the picture isn't a cone. that's a circular sector that was used to produce a cone. after drawing the ellipse around the cone, the cone was opened to the parent sector. the ellipse's perimeter looks like what you see in the picture. –  Mar 08 '20 at 18:17
  • @epimorphic after this, I made $3$ more sectors to draw ellipse on the cone. then I put the $4$ opened perimeters together along and the result was something like what you can see in $x^2+xy+y^2=a^2$ graph. ($a$ can be any number) –  Mar 08 '20 at 18:20
  • I don't quite follow your last reply. What I'm saying is that in your picture the angle and the 6cm length specify the cone (the height and the radius) you get once you roll it up, and the 3cm measurement gives you a single point of the ellipse on the cone (up to rotation). But if only one point is fixed you can tilt the plane on that point in various ways to produce ellipses of various axes and perimeters. – epimorphic Mar 08 '20 at 18:33
  • An example of a solvable situation is if you know the distances of the closest and the farthest points of the ellipse from the cone vertex. – epimorphic Mar 08 '20 at 18:39
  • I fortunately have another ellipse and I know the distances of the closest and the farthest points of it from the cone's vertex. the closest point has a distance of $\frac{h^4}{l^3}$ and the farthest point has a distance of $\frac{h^2}{l}$ from the cone vertex. (h is cone's height and l is the cone's lateral side). for example when $h=100$ and $l=125$. If you want to know how these distances were computed, it was by using Pythagorean rule, Thales' theorem and Trigonometry. trust me; they are accurate. @epimorphic –  Mar 08 '20 at 19:15
  • in the picture, yes. the left and right points make a single point for the ellipse on the cone –  Mar 08 '20 at 19:15
  • @aminabzz, many such ellipses are possible passing through a point on the cone at a distance of $3$ from the center. Also, you need the line along which the cone is being cut to net it into a sector. – SarGe Aug 25 '20 at 07:22

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