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# Write down parametric equation for the conic: (5\4)*x^2 + 3y^2 + 5x -6y - 37 = 0

vagyika | (Level 1) Valedictorian

Posted October 29, 2013 at 4:52 PM via web

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Write down parametric equation for the conic: (5\4)*x^2 + 3y^2 + 5x -6y - 37 = 0

justaguide | College Teacher | (Level 2) Distinguished Educator

Posted October 29, 2013 at 5:26 PM (Answer #1)

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The equation of the ellipse is `(5\4)x^2 + 3y^2 + 5x -6y - 37 = 0`

`(5\4)x^2 + 3y^2 + 5x - 6y - 37 = 0`

=> `5x^2 + 12y^2 + 20x - 24y - 148 = 0`

=> `5(x^2 + 4x + 4) + 12y^2 - 24y - 148 - 20 = 0`

=> `5(x^2 + 4x + 4) + 12(y^2 - 2y + 1) - 148 - 20 - 12 = 0`

=> `5(x + 2)^2 + 12(y - 1)^2 - 180 = 0`

=> `5(x + 2)^2 + 12(y - 1)^2 = 180`

=> `(x + 2)^2/36 + (y - 1)^2/21 = 1`

=> `(x + 2)^2/6^2 + (y - 1)^2/(sqrt 21)^2 = 1`

The parametric form of this ellipse is `x = -2 + 6*cos t ` and `y = 1 + sqrt 21*sin t`

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justaguide | College Teacher | (Level 2) Distinguished Educator

Posted October 30, 2013 at 4:04 AM (Reply #1)

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Sorry for a small typo in the answer.

The equation in the standard form is:

`(x +2)^2/36 + (y - 1)^2/15 = 1`

=> `(x +2)^2/6^2 + (y - 1)^2/(sqrt 15)^2 = 1`

a = 6 and b =

The parametric form is `x = -2 + 6*cos t` and `y = 1 + sqrt 15*sin t`

vagyika | (Level 1) Valedictorian

Posted October 29, 2013 at 9:17 PM (Answer #2)

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You had b= root 21, isn't it root 15 ??

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