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Math Help - Ecentricity

  1. #1
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    Exclamation Ecentricity

    How do you find the eccentricity of 4x^2 + 16y^2 + 8x +64y +4 = 0
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  2. #2
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    Quote Originally Posted by jumpman23 View Post
    How do you find the eccentricity of 4x^2 + 16y^2 + 8x +64y +4 = 0
    1. By completing the square you get:

    4x^2 + 16y^2 + 8x +64y +4 = 0

    4(x^2+2x\bold{\color{blue}+1}) + 16(y^2  +4y \bold{\color{red}+4}) = -4\bold{\color{blue}+4}\bold{\color{red}+64}

    \dfrac{(x+1)^2}{16}+\dfrac{(y+2)^2}{4}=1

    2. You are supposed to know that

    e^2+b^2=a^2 Plug in the values you know and calculate e:

    e^2=16-4=12~\implies~\boxed{e=2\sqrt{3}}
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  3. #3
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    thanks
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  4. #4
    Senior Member pankaj's Avatar
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    Sorry, earboth the answer is wrong.The curve in question is an ellilse whose eccentricity must be less than 1.

    The formula used by you for calculating eccentricity is wrong.It should be

     <br />
a^2=b^2+a^2e^2<br />

    Thus e=\frac{\sqrt{3}}{2}
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  5. #5
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    Quote Originally Posted by pankaj View Post
    Sorry, earboth the answer is wrong.The curve in question is an ellilse whose eccentricity must be less than 1.

    The formula used by you for calculating eccentricity is wrong.It should be

     <br />
a^2=b^2+a^2e^2<br />

    Thus e=\frac{\sqrt{3}}{2}
    Thanks for your reply - but in my opinion:

    1. there exist a linear eccentricity e^2=a^2-b^2. That's the eccentricity I've calculated.

    2. there exist a numerical eccntricity \dfrac ea=\epsilon. Obviously this is the eccentricity you've had in mind.

    (Remark: I translated the technical terms literally from German so maybe the English expressions differ a bit)
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  6. #6
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    Talking

    To confirm the formula they gave you in your book for the eccentricity of an ellipse, and for worked examples of how to find the eccentricity, try some online lessons.

    Then make sure that you memorize the formula before your next test!!
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  7. #7
    Senior Member pankaj's Avatar
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    Sorry.Actually I had never read about Linear and numerical eccentricity
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  8. #8
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    Quote Originally Posted by earboth View Post
    Thanks for your reply - but in my opinion:

    1. there exist a linear eccentricity e^2=a^2-b^2. That's the eccentricity I've calculated.

    2. there exist a numerical eccntricity \dfrac ea=\epsilon. Obviously this is the eccentricity you've had in mind.

    (Remark: I translated the technical terms literally from German so maybe the English expressions differ a bit)
    What you calculated would be called, in English, the "focal length"- the distance from the center of the ellipse to each focus.
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