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Math Help - help finding foci

  1. #1
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    Question help finding foci

    x^2+8y^2+6x-16x+1=0

    x^2+6x+9 +8(y^2-2y+1)=-1+9+1=9

    (x+3)^2/9 + 8(y-1)^2/9

    h=-3

    i am supposed to get sqrt 14 but i just dont see how. Can somebody please explain to me why it is supposed to be sqrt 14? Thank You.
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  2. #2
    A riddle wrapped in an enigma
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    Quote Originally Posted by blackhawk9876 View Post
    x^2+8y^2+6x-16x+1=0

    x^2+6x+9 +8(y^2-2y+1)=-1+9+1=9

    (x+3)^2/9 + 8(y-1)^2/9

    h=-3

    i am supposed to get sqrt 14 but i just dont see how. Can somebody please explain to me why it is supposed to be sqrt 14? Thank You.
    x^2+6x+9+8(y^2-2y+1)=-1+9+8

    (x+3)^2+8(y-1)^2=16

    \frac{(x+3)^2}{16}+\frac{(y-1)^2}{2}=1

    center \ \ = \ \ (h, k) \ \ = \ \ (-3, 1)

    a^2 = 16
    a=4

    b^2 = 2
    b = \sqrt2

    c^2=a^2-b^2

    Can you finish?

    The Following is copied from Jhevon's referenced post. This should finish things up nicely. Thanks Jhevon!!

    <Quote=Jhevon>

    The equation of an ellipse with a horizontal major axis can be expressed as

    <br />
\frac {(x - h)^2}{a^2} + \frac {(y - k)^2}{b^2} = 1 \mbox { for } \boxed {a \geq b > 0}<br />

    When in this form, the following hold:

    Center: (h,k)
    Vertices: (h \pm a,k)
    Foci: (h \pm c, k)
    ........where c^2 = a^2 - b^2

    <End Quote = Jhevon>
    Last edited by masters; June 25th 2008 at 05:45 PM.
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  3. #3
    is up to his old tricks again! Jhevon's Avatar
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    For reference, see post #2 here
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  4. #4
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    Yes I can finish, thank you. I kept adding 1 instead of 8 thank you. Funny how you can get so stumped and its just a simple error messing the whole problem up
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  5. #5
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by blackhawk9876 View Post
    Yes I can finish, thank you. I kept adding 1 instead of 8 thank you. Funny how you can get so stumped and its just a simple error messing the whole problem up
    yeah, hehe, it's funny afterwards ...frustrating during
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