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Simple Conic I am sure - Overlooking it

  1. Sep 26, 2007 #1
    1. The problem statement, all variables and given/known data
    Graph 4x^2 + 9y^2=1

    3. The attempt at a solution
    I have no idea where to start. There is nothing to divide out by, and there are no A or B terms. Any assistance is greatly appriciated!
    Thanks,
    Chris
     
  2. jcsd
  3. Sep 26, 2007 #2

    EnumaElish

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    Hint: what would you get with x^2 + y^2 = 1? Different equation, same idea.
     
  4. Sep 26, 2007 #3
    A circle with radius 1. So does this mean I get an ellipse, centered at the origin with a=2 and b=3?
     
  5. Sep 26, 2007 #4

    arildno

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    Hint:
    Make use of the identity
    [tex]4x^{2}=(\frac{x}{\frac{1}{2}})^{2}[/tex]
     
  6. Sep 26, 2007 #5

    symbolipoint

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    It is an ellipse with center at the origin. Do you know how to use the standard form of the equation of an ellipse to find the vertical and horizontal lengths of the ellipse? Intermediate Algebra; standard form for untranslated ellipse is:
    (x^2)/(a^) + (y^2)/(b^2) = 1;
    What do the "a" and the "b" tell you?
     
  7. Sep 26, 2007 #6
    A and B tell you the distance from the center of the ellipse to make a point. A tells you how far in the verticle direction (up and down) to move, and B tells horizontal direction to move (left and right). I don't understand this however becuase there are no A and B, unless both are 1. I don't know why I am getting so hung up on this problem, I did all my other conics fine.
     
  8. Sep 26, 2007 #7

    symbolipoint

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    Excuse me for omitting one of the symbols in the standard ellipse equation. I meant to write: (x^2)/(a^2) + (y^2)/(b^2) = 1
     
  9. Sep 26, 2007 #8
    Rewrite the 4 so that you have x^2/something. Do the same thing with the 3
     
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