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Math Help - eccentrity of hyperbola

  1. #1
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    eccentrity of hyperbola

    find the eccentricity of the hyperbola.

    x^2-2y^2=6

    I got 3sqrt of 6/ 6
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  2. #2
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    Quote Originally Posted by scottydoint View Post
    find the eccentricity of the hyperbola.

    x^2-2y^2=6

    I got 3sqrt of 6/ 6
    For the hyperbola \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1 the eccentricity is given by e = \sqrt{1 + \frac{b^2}{a^2}}.

    For your hyperbola a = \sqrt{6} and b = \sqrt{3}. Therefore e = \frac{\sqrt{6}}{2} which is what your answer simplifies to.

    So yes you are correct.
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  3. #3
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    two graphs

    Sorry for all of these questions, but this is last minute and my teacher can't even get these problems right in class so I am trying to teach myself from a book that is hard to follow.

    How do you graph this parabola:
    x=-3y^2-6y+3

    and how do you graph this circle? I don't think its possible.

    x^2+(y-1)^2+1=0
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  4. #4
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    Quote Originally Posted by scottydoint View Post
    Sorry for all of these questions, but this is last minute and my teacher can't even get these problems right in class so I am trying to teach myself from a book that is hard to follow.

    How do you graph this parabola:
    x=-3y^2-6y+3

    Mr F says: The standard form is found by completing the square:

    {\color{red}x = -3 (y^2 + 2y - 1) = -3 [(y+1)^2 - 2] = -3(y+1)^2 + 6} \, ....



    and how do you graph this circle? I don't think its possible.

    x^2+(y-1)^2+1=0 Mr F says: {\color{red} \Rightarrow x^2 + (y-1)^2 = -1} which is clearly not possible for real values of x and y.
    New questions should be asked in a new thread.
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  5. #5
    Rhymes with Orange Chris L T521's Avatar
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    Quote Originally Posted by mr fantastic View Post
    New questions should be asked in a new thread.
    I had already answered the question here...

    --Chris
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  6. #6
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    Sorry about that. I posted it in this thread thinking I was posting a new one and realized after I did it and posted a new one.
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  7. #7
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    Quote Originally Posted by Chris L T521 View Post
    I had already answered the question here...

    --Chris
    Our replies are sufficiently different that they more or less complement each other. I've linked this thread to the other.
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