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Find the length of the Transverse axis. What did I do wrong?

  1. Mar 8, 2005 #1
    I'm doing an Exam correction an I can't see how this answer is found.

    Question: [tex]x^2 - \frac{\ y^2}{9} = 1[/tex]
    Find the length of the transverse axis.

    I took [tex]x^2 - \frac{\ y^2}{9} = 1[/tex]

    And plugged it into
    [tex]\frac{\ (x-h)^2}{b^2} - \frac{\ (y-k)^2}{a^2} = 1[/tex]

    to get

    [tex]\frac{\ (x-0)^2}{1^2} - \frac{\ (y-0)^2}{3^2} = 1[/tex]

    The 0's were because in the original equation, I didn't see an [tex]h[/tex] or [tex]k[/tex] value that was affecting either [tex]x[/tex] or [tex]y[/tex]
    The [tex]1^2[/tex] is because [tex]x[/tex] = [tex]\frac{x}{1}[/tex] like [tex]a[/tex] = [tex]\frac{a}{1}[/tex]

    [tex]a = 3[/tex], therefore the length of the transverse is 6 units...



    ....Incorrect Answer??!
     
  2. jcsd
  3. Mar 8, 2005 #2

    HallsofIvy

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    Staff Emeritus
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    When x= 0 what is y?

    Setting x= 0 gives [itex]-\frac{y^2}{9}= 1[/itex]- which has NO solution.

    When y= 0 what is x?

    Setting y= 0 gives x2= 1 so x= -1 and 1. THAT'S the "transverse axis" you want!
     
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