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Math Help - Finding the foci of an ellipse from an equation.

  1. #1
    Member MathBlaster47's Avatar
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    Finding the foci of an ellipse from an equation.

    I have an odd question, that I think I have worked out properly, but the answer makes no sense.

    What are (a) the x-intercepts,(b) the y-intercepts and(c) the foci of the ellipse \frac{x^2}{25}+ \frac{y^2}{36}=1.

    I have already determined that the x and y intercepts are: \pm5 and \pm6,respectively, meaning that the major axis is vertical as the y-intercept is greater than the x-intercept.
    My problem is that when I use the relation: c^2=a^2-b^2
    the result is -9, so c is the square root of -9--which doesn't exist and cant be graphed, meaning I did something wrong.....
    What was it that I did wrong?
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  2. #2
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    Quote Originally Posted by MathBlaster47 View Post
    I have an odd question, that I think I have worked out properly, but the answer makes no sense.

    What are (a) the x-intercepts,(b) the y-intercepts and(c) the foci of the ellipse \frac{x^2}{25}+ \frac{y^2}{36}=1.

    I have already determined that the x and y intercepts are: \pm5 and \pm6,respectively, meaning that the major axis is vertical as the y-intercept is greater than the x-intercept.
    My problem is that when I use the relation: c^2=a^2-b^2
    the result is -9, so c is the square root of -9--which doesn't exist and cant be graphed, meaning I did something wrong.....
    What was it that I did wrong?
    the foci are on the y-axis ... c^2 = b^2-a^2

    also, check you arithmetic when calculating ... (it's not 9)
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  3. #3
    Member MathBlaster47's Avatar
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    Quote Originally Posted by skeeter View Post
    the foci are on the y-axis ... c^2 = b^2-a^2

    also, check you arithmetic when calculating ... (it's not 9)
    GYAHH....(insert expletive), 1+1=2 and 36-25=11.....typing too fast is killing me....Must lay off the sugar.....
    Ok! I think I understand what is going on, since the axis is on y then the equation for establishing the foci changes to reflect that, meaning that I have to turn the numbers around.
    Meaning that the foci are (0,\sqrt{11}) and (0,-\sqrt{11})?
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  4. #4
    Junior Member StonerPenguin's Avatar
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    Quote Originally Posted by MathBlaster47 View Post
    GYAHH....(insert expletive), 1+1=2 and 36-25=11.....typing too fast is killing me....Must lay off the sugar.....
    Ok! I think I understand what is going on, since the axis is on y then the equation for establishing the foci changes to reflect that, meaning that I have to turn the numbers around.
    Meaning that the foci are (0,\sqrt{11}) and (0,-\sqrt{11})?
    Yup, that's what I got;
    The x-intercepts are 5 and -5
    The y-intercepts are 6 and -6
    And the foci of the ellipse are (0,\sqrt{11}) and (0,-\sqrt{11})
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