Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

An ellipse question?

  1. Feb 19, 2007 #1
    i added a file with the curve the is being asked to find

    an ellipse is given. its formula is x^2 + 2*y^2=8 .

    find the formula of the curve that is being created by the

    the centers of the perpendicular lines to the X axes and the ellipse.


    i tried to solve this question
    by putting the Y^2 on the one side and on the other the rest
    and devidind it by 2 .


    2y^2=8-x^2
    y^2=8-x^2/2
    y=V(8-x^2/2)
    (v=root simbol)

    the curve that in the center always smaller in height by 2
    so i devided the formula by 2 to find our curve

    y=1/2*V(8-x^2/2)
    in my book it gives a different answer

    (x^2)/8 +y^2=1

    plz help
     

    Attached Files:

    • el4.JPG
      el4.JPG
      File size:
      6.5 KB
      Views:
      143
  2. jcsd
  3. Feb 19, 2007 #2

    HallsofIvy

    User Avatar
    Science Advisor

    What do you mean by "the centers of the perpendicular lines to the X axes and the ellipse"?
     
  4. Feb 19, 2007 #3
    like it shows at the pictue
    a curve wich composed from the middle points of this straight lines
    parrallel to the Y axes
     
  5. Feb 20, 2007 #4
    how can i solve this thing???
     
  6. Feb 20, 2007 #5

    HallsofIvy

    User Avatar
    Science Advisor

    [itex]x^2+ 2y^2= 8[/itex] is the same as [itex]\frac{x^2}{8}+ \frac{y^2}{4}= 1[/itex].
    That's an ellipse with major semi-axis, along the x-axis, of length [itex]\sqrt{8}= 2\sqrt{2}[/itex] and minor semi-axis, along the y-axis, of length 2.

    Dividing the y-coordinate of each point by 2 gives an ellipse with the same major semi-axis but minor semi-axis of length 1:
    [itex]\frac{x^2}{8}+ y^2=1[/itex].
     
  7. Feb 20, 2007 #6

    HallsofIvy

    User Avatar
    Science Advisor

    Here's your error! Don't know why I didn't see this sooner. Dividing both sides of 2y2= 8- x2 by 2 gives y2= 4- x2!

    [itex]y= 1/2 \sqrt{4- x^2}[/itex]

    Square both sides of [itex]2y= \sqrt{4- x^2}[/itex] and you get
    [itex]4y^2= 4- x^2[/itex] or [itex]x^2+ 4y^2= 4[/itex].

    Divide through by 4:
    [tex]\frac{x^2}{4}+ y^2= 1[/itex]
     
  8. Feb 21, 2007 #7
    thank you very much
     
  9. Feb 22, 2007 #8

    HallsofIvy

    User Avatar
    Science Advisor

    Now, my question is, "what does this have to do with 'Tensor Analysis and Differential Geometry'?"
     
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook