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Math Help - Homework help please!

  1. #1
    Junior Member rednest's Avatar
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    Angry Homework help please!

    Question:

    The point P(7 cos t,5 sin t) is on the ellipse with equation \frac{x^{2}}{49} + \frac{y^{2}}{25} = 1. The line through P parallel to the y-axis meets the x-axis at X. The point Q is on the line XP produced so that XQ = 2XP. Find, in cartesian form, an equation of the locus of Q, as t varies.
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  2. #2
    Member SengNee's Avatar
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    Pangkor Island, Perak, Malaysia.
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    Quote Originally Posted by rednest View Post
    Question:

    The point P(7 cos t,5 sin t) is on the ellipse with equation \frac{x^{2}}{49} + \frac{y^{2}}{25} = 1. The line through P parallel to the y-axis meets the x-axis at X. The point Q is on the line XP produced so that XQ = 2XP. Find, in cartesian form, an equation of the locus of Q, as t varies.
    P(7cost,5sint)
    X(7cost,0)
    Q(7cost,y)

    \frac{x^{2}}{49} + \frac{y^{2}}{25} = 1
    \frac{(7cost)^{2}}{49} + \frac{(5sint)^{2}}{25} = 1
    cos^2t+sin^2t= 1

    XQ=2XP
    XQ^2=4XP^2
    (7cost-7cost)^2+(y-0)^2=4[(7cost-7cost)^2+(5sint-0)^2]
    y^2=100sin^2t
    y=10sint

    What is cartesian form?
    Is it Q(7cost,10sint)?
    My answer is correct?
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  3. #3
    Super Member

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    Lexington, MA (USA)
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    Hello, rednest!

    The point P(7\cos\theta,5\sin\theta) is on the ellipse with equation: \frac{x^{2}}{49} + \frac{y^{2}}{25} = 1

    The line through P parallel to the y-axis meets the x-axis at X

    The point Q is on the line XP produced so that XQ = 2XP

    Find, in cartesian form, an equation of the locus of Q, as \theta varies.
    Code:
                        |
                      * * *
                *       |       *   P
             *          |          o      -
           *            |       *  | *    :
                        |     *    |      : y
          *             |  * θ     |  *   :
      ----*-------------+----------+--*----
          *             |         X|  *   :
                        |          |      :
           *            |          | *    :
             *          |          *      : 2y
                *       |       *  |      :
                      * * *        |      :
                        |          |      :
                        |          |      :
                        |          o      -
                        |          Q

    After all that sketching and a lot of thinking, it turns out that

    . . the equations for Q are: . \begin{Bmatrix}x &=& 7\cos\theta \\ y &=&\text{-}10\sin\theta \end{Bmatrix}


    Eliminate the parameter: . \begin{Bmatrix}\dfrac{x}{7} &=&\cos\theta \\  \\[-1mm]\text{-}\dfrac{y}{10} &=&\sin\theta \end{Bmatrix}

    . . . . . . . . . . . . Square: . \begin{Bmatrix}\dfrac{x^2}{49} &=& \cos^2\!\theta \\ \\[-1mm] \dfrac{y^2}{100} &=&\sin\theta \end{Bmatrix}

    \text{Add: }\;\;\frac{x^2}{49} + \frac{y^2}{100} \;=\;\underbrace{\cos^2\!\theta + \sin^2\!\theta}_{\text{This is 1}}


    Therefore: . \boxed{\frac{x^2}{49} + \frac{y^2}{100} \;=\;1}

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