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Math Help - Ellipses

  1. #1
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    Ellipses

    I get the general idea of finding the Verticies, Foci, Eccentricity, Major Axes, and the Minon Axes.

    However i'm having trouble doing it with this problem

    2x^2 + y^2 = 3


    Can someone point me in the right direction? I know I need to get it to equal 1 so I divide the entire equation by 3. This in turn will give me

    (2/3)x^2 + (y^2)/ 3 = 1

    Well looking at the numbers, x has a greater value than y since it is 2/3. So the verticies should be (+-Sqrt(2/3),0) but when I check the back of the book, it's not giving me that answer. It gives me (0,+-sqrt(3))

    Help
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  2. #2
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    Another one i'm having problems with is

    x^2 + 4y^2 = 1
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  3. #3
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by JonathanEyoon View Post
    I get the general idea of finding the Verticies, Foci, Eccentricity, Major Axes, and the Minon Axes.

    However i'm having trouble doing it with this problem

    2x^2 + y^2 = 3
    Yes, you need to start by dividing by 3. But you need to put it into the correct form:
    2x^2 + y^2 = 3

    \frac{2}{3}x^2 + \frac{1}{3}y^2 = 1

    \frac{x^2}{\frac{3}{2}} + \frac{y^2}{3} = 1

    \frac{x^2}{\left ( \sqrt{ \frac{3}{2}} \right ) ^2} + \frac{y^2}{(\sqrt{3})^2} = 1

    So a = \sqrt{3} is the semimajor axis and b = \sqrt{\frac{3}{2}} is the semiminor axis.

    Does that help you out?

    -Dan
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  4. #4
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    That helps me out quite a bit




    Quote Originally Posted by topsquark View Post
    Yes, you need to start by dividing by 3. But you need to put it into the correct form:
    2x^2 + y^2 = 3

    \frac{2}{3}x^2 + \frac{1}{3}y^2 = 1

    \frac{x^2}{\frac{3}{2}} + \frac{y^2}{3} = 1

    \frac{x^2}{\left ( \sqrt{ \frac{3}{2}} \right ) ^2} + \frac{y^2}{(\sqrt{3})^2} = 1

    So a = \sqrt{3} is the semimajor axis and b = \sqrt{\frac{3}{2}} is the semiminor axis.

    Does that help you out?



    -Dan
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  5. #5
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    So the second one I posted I'm having a problem with should be?
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  6. #6
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by JonathanEyoon View Post
    So the second one I posted I'm having a problem with should be?
    x^2 + 4y^2 = 1

    \frac{x^2}{1^2} + \frac{y^2}{ \left ( \frac{1}{2} \right ) ^2} = 1

    -Dan
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  7. #7
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    Quote Originally Posted by topsquark View Post
    x^2 + 4y^2 = 1

    \frac{x^2}{1^2} + \frac{y^2}{ \left ( \frac{1}{2} \right ) ^2} = 1

    -Dan

    For the denominator in Y^2, how did you get 1/2?
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  8. #8
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by JonathanEyoon View Post
    For the denominator in Y^2, how did you get 1/2?
    Step by step, then.

    x^2 + 4y^2 = 1

    \frac{x^2}{1} + \frac{y^2}{\frac{1}{4}} = 1

    \frac{x^2}{1^2} + \frac{y^2}{\left ( \frac{1}{2} \right ) ^2} = 1

    -Dan
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