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Vertical or Horizontal ellipse?

  1. Apr 2, 2005 #1
    [tex] 16x^2+9y^2+192y-36y+468=0 [/tex]

    Was the original conic i had to conver this into standard form and got

    [tex] \frac {(x+6)^2} {9} + \frac {(y-2)^2} {16} =1 [/tex]

    Im not sure if this is a horizontal or vertical ellipse
     
  2. jcsd
  3. Apr 2, 2005 #2
    the center of the ellipse is [itex](-6, 2)[/itex]. If you get [itex]x=-6[/itex], then what range of values can [itex]y[/itex] take? If you set [itex]y = 2[/itex] then what range of values can [itex]x[/itex] take?

    Can these facts help you to decide? If you can't see why they can directly, then draw a picture of the ellipse and see if you can tell~
     
  4. Apr 2, 2005 #3
    since the denominator of the x is less that the denominator of the y then the equation is in the form

    [tex] \frac {(x-h)^2} {b^2} + \frac {(y-h)^2} {a^2} [/tex]

    I think if what I said is right then this is a vertical ellipse

    a=4 b=3?
     
    Last edited: Apr 2, 2005
  5. Apr 2, 2005 #4
    indeed it is.
     
  6. Apr 2, 2005 #5
    Here are the features i got for this conic
    Vertical ellipse
    Center (-6,2) a=4 b=3
    Length of major axis 2a=8
    Length of minor axis 2b=6
    Vertices (-6,-2) and (-6,6)
    Foci=(-6,-3) and (-6,7) where c=5

    are all of these correct?
    :smile:
     
    Last edited: Apr 2, 2005
  7. Apr 3, 2005 #6
    The focus is outsie of the ellipse? What is wrong? Is this ok?
     
  8. Apr 3, 2005 #7
    (h,-c+k) and (h,c+k) are the foci because this is a vertical ellipse

    c=5 and I plugged in the center (-6,2) wats wrong?
     
  9. Apr 3, 2005 #8
    what's [itex]c[/itex], and why do you think it's [itex]5[/itex]?

    If you're using it that way, then it should be [itex]\sqrt{7}[/itex] (I made a mistake earlier, by the way... that's why the other post is gone now :wink:)
     
    Last edited: Apr 3, 2005
  10. Apr 3, 2005 #9
    c= sqrt(a^2+b^2) how is it sqrt7?
     
  11. Apr 3, 2005 #10
    it's actually

    [tex]b^2 = a^2 - c^2 \Longrightarrow c^2 = a^2 - b^2 \Longrightarrow c = \sqrt{a^2-b^2}[/tex]

    using your formula, you would always find the focii outside the ellipse.
     
  12. Apr 4, 2005 #11
    Thanks soo much I made the same mistake in 3 problems now I rememeber thankssssssss LIFE SAVER!! :smile:
     
  13. Apr 4, 2005 #12
    If the value under y^2 is greater, then it's going to be a vertical ellipse.
    If the value under x^2 is greater, then it's going to be horizontal.

    You can check by graphing and calculating the lengths of the major and minor axes. That should help you too.
     
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