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Math Help - Conics: ellipse in standard position

  1. #1
    Junior Member froodles01's Avatar
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    Conics: ellipse in standard position

    An example with answer I have goes;
    x = 6 cos t
    y = sqrt 11 sin t

    Write down the equation of the conic & state what type of conic it is.

    trig identity cos2t + sin2t = 1

    which gives

    x2 / 62 + y2 / (sqrt 11)2 = 1 . . . . . . . .

    er! how does this happen, if you could help!
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  2. #2
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    Re: Conics: ellipse in standard position

    x = 6\cos(t), \ y = \sqrt{11}\sin(t), so

    \cos(t) = \frac{x}{6}, \ \sin(t) = \frac{y}{\sqrt{11}}, so

    \cos^2(t) = \left(\frac{x}{6}\right)^2, \ \sin^2(t) = \left(\frac{y}{\sqrt{11}}\right)^2, so

    \cos^2(t) = \frac{x^2}{6^2}, \ \sin^2(t) = \frac{y^2}{(\sqrt{11})^2}.

    NOW, use the trigonometry formula \cos^2(t) + \sin^2(t) = 1, and plug in what we just found.

    \cos^2(t) + \sin^2(t) = 1, so

    \left( \frac{x^2}{6^2}\right) + \left( \frac{y^2}{(\sqrt{11})^2} \right) = 1, so

    \frac{x^2}{6^2} + \frac{y^2}{(\sqrt{11})^2} = 1.

    You could rewrite that as:

    \frac{x^2}{36} + \frac{y^2}{11} = 1.

    Did that help?
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  3. #3
    Junior Member froodles01's Avatar
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    Re: Conics: ellipse in standard position

    A simple answer for a simple problem - thank you for taking the time. :-)
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