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Finding an ellipse from a plane slicing a cone

  1. Jul 29, 2011 #1
    Hello,

    So I'm doing some independent study and I'm at a loss for this problem.

    1. The problem statement, all variables and given/known data

    Let's say we have an ellipse of the form (x2)/a + (y2)/b = 1 which we obtain by slicing a plane through a right circular cone with an opening angle of [itex]\theta[/itex] (a fixed constant). We know the angle the plane makes with the xy-plane to be [itex]\phi[/itex]. We also know the cut crosses the center line of the cone at height h. In terms of h and [itex]\phi[/itex], what is a and b?

    2. Relevant equations

    I found this page that seemed to be pointing me in the right direction, but I can't make much sense of it:

    http://www.physicsinsights.org/conic_sections_1.html

    Also, the equation of the ellipse they are using is related to the foci of the ellipse which I don't know how to relate to the equation of an ellipse I'm looking for.

    3. The attempt at a solution

    I'm honestly not sure where to start or if this is possible. It seems to me like it should be, so any help would be greatly appreciated.

    Thank you,
    -anomie
     
  2. jcsd
  3. Jul 29, 2011 #2

    tiny-tim

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    Last edited by a moderator: May 5, 2017
  4. Jul 29, 2011 #3
    Thanks for the quick reply!

    I'm imagining the process to get the major and minor radius of the ellipse would go something like this: Get the points of each sphere that touch the plane from above and below. This will be the foci of your ellipse, which in turn you can formulate the equation of the ellipse.

    So how would I go about getting the Dandelin spheres with just the information in my above post? Essentially, how do I get the Dandelin spheres tangent to both a given plane and a given cone?

    Thank you,
    -anomie
     
  5. Jul 30, 2011 #4

    tiny-tim

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    hello anomie! :smile:

    (just got up :zzz: …)
    you wouldn't bother … there obviously are such spheres (they're the largest possible inscribed spheres), and their mere existence proves that it is an ellipse (or parabola or hyperbola, depending on the slope)

    once you know it is an ellipse, you know that the minor semi-axis, √b, is simply the radius of the cone at height h, and you can find the major semi-axis, √a, from the obvious triangle :wink:

    (and you can find the foci using the usual formula)

    btw, we usually write an ellipse with a and b squared … x2/a2 + y2/b2 = 1 … to make calculations easier!)
     
  6. Jul 30, 2011 #5
    Ok, sweet, things are starting to make more sense. Thanks for the great help!:smile:

    I was easily able to get the minor axis. As for the major axis, I'm not getting the right results. I tried using the right triangle that is formed on the ellipse with the sides being the semi-minor and semi-major axes. I determined that the angle formed with the hypotenuse and semi-major axis is always going to be half of opening angle ([itex]\theta[/itex]/2). Thus, knowing the semi-minor axis, I can solve for a using a=b/(tan([itex]\theta[/itex]/2)). An image below to show what I mean:

    rQviq.png

    The thing that I'm not sure on is that the angle is indeed always half of the opening angle (which I'm really starting to doubt now). The other triangle I tried to use was the one formed by the center line of the cone, the semi-major axis, and the line formed by the edge of the cone. Yet, I still get vastly disproportional results.

    Thank you,
    -anomie
     
  7. Jul 31, 2011 #6

    tiny-tim

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    hi anomie! :smile:
    no it isn't! :redface:
    that should do it …

    what result did you get? :smile:
     
  8. Jul 31, 2011 #7
    Hello!

    So it looks like I was doing it right overall but just had some arithmetic errors I glanced over when it came to the finding the major axis. The math makes sense to me know thanks to you!:smile:

    Thanks a lot for the help!

    -anomie
     
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