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Math Help - Parametric equation of an ellipse

  1. #1
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    Parametric equation of an ellipse

    Can somebody please help me if the following is correct:

    I must give parametric equations for the following ellipses:

    1) 3x^2 + 5y^2 = 75

    Since a=5 and b=sqrt(15), is this my correct answer:
    x = 5 cos t
    y = sqrt(15) sin t

    2) 3x^2 + 12x + 5y^2 -10y -58 =0

    Since a=5 and b=sqrt(15) like above mentioned, but this ellipse is transformed so that the new center became (-2, 1), then is it that the parametric equations became:
    x= -2 + (5 cos t)
    y= 1 + (sqrt(15) sin t)

    Many thanks to someone who helps!
    Brigitte
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  2. #2
    Senior Member Stroodle's Avatar
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    I got the same answers, so your answers look correct to me
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  3. #3
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    Quote Originally Posted by dolkam View Post
    Can somebody please help me if the following is correct:

    I must give parametric equations for the following ellipses:

    1) 3x^2 + 5y^2 = 75

    Since a=5 and b=sqrt(15), is this my correct answer:
    x = 5 cos t
    y = sqrt(15) sin t

    2) 3x^2 + 12x + 5y^2 -10y -58 =0

    Since a=5 and b=sqrt(15) like above mentioned, but this ellipse is transformed so that the new center became (-2, 1), then is it that the parametric equations became:
    x= -2 + (5 cos t)
    y= 1 + (sqrt(15) sin t)

    Many thanks to someone who helps!
    Brigitte
    You should be checking your answer by substituting the parametric equation into the cartesian ones. They are both correct.

     3x^2 + 5y^2 = 75
    Subbing in  x = 5 \cos t and  y = \sqrt{15} \sin t
     = 75 \cos^2 t + 75 sin^2 t
     = 75 (cos^2 t + sin^2 t) = 75


     3x^2 + 12x + 5y^2 -10y -58 =0
     3(x + 2)^2 + 5(y-1)^2 = 58 + 12 + 5
    Subbing in  x = -2 + 5 \cos t and  y = 1 + \sqrt{15} \sin t
     = 75 \cos^2 t + 75 sin^2 t
     = 75 (cos^2 t + sin^2 t) = 75
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  4. #4
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    Thank you very much.

    I did not think about the substituting into the cartesian.

    I will do that.

    Brigitte
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