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Math Help - Parametric Equations

  1. #1
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    Parametric Equations

    I'm having problems with a few parametric equation questions. Considering how much trouble I've been having with this particular unit of work, I expect I'll have to use this thread to ask more questions in the near future. The one I'm stuck on at the moment is;

    PQ, a variable chord of the parabola x^2 = 4y, subtends a right angle at the vertex O.

    a) If p and q are parameters corresponding to P and Q, show that pq = -4.

    b) If the tangents at P and Q intersect at T, find the locus of T.

    c) If M is the midpoint of chord PQ, determine the locus of M.
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  2. #2
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    Hello, Flay!

    PQ, a variable chord of the parabola x^2 \,= \,4y, subtends a right angle at the vertex O.

    a) If p\text{ and }q are parameters corresponding to P and Q, show that: pq = -4.
    Parametric equations: . \begin{array}{ccc}x &=& 2t \\ y &=&t^2 \end{array}

    The points are: . \begin{array}{c}P\,(2p,p^2) \\ Q\,(2q,q^2) \end{array}

    The slope of OP is: . m_{OP} \:=\:\frac{p^2 - 0}{2p-0} \:=\:\frac{p}{2}

    The slope of OQ is: . m_{OQ} \:=\:\frac{q^2-0}{2q-0} \:=\:\frac{q}{2}

    Since OP \perp OQ, we have: . \frac{p}{2} \:=\:-\frac{2}{q}\quad\Rightarrow\quad\boxed{ pq \:=\:-4}


    Note: .Since q = -\frac{4}{p}, point Q becomes: . \left(-\frac{8}{p},\:\frac{16}{p^2}\right)




    b) If the tangents at P\text{ and }Q intersect at T, find the locus of T.
    The slope of the tangent is: . \frac{dy}{dx} \:=\:\frac{\frac{dy}{dt}}{\frac{dx}{dt}} \:=\:\frac{2t}{2} \:=\:t


    At P, we have the point (2p, p^2) and slope p.
    The equation of the tangent at P is:
    . . y - p^2 \:=\:p(x-2p)\quad\Rightarrow\quad y \:=\:px - p^2\;\;{\color{blue}[1]}

    At Q, we have the point \left(\text{-}\frac{8}{p},\:\frac{16}{p^2}\right) and slope -\frac{4}{p}
    The equation of the tangent at Q is:
    . . y - \frac{16}{p^2} \:=\:-\frac{4}{p}\left(x + \frac{8}{p}\right)\quad\Rightarrow\quad y \:=\:-\frac{4}{p}x - \frac{16}{p^2}\;\;{\color{blue}[2]}

    The two tangents intersect: . px-p^2 \:=\:-\frac{4}{p}x - \frac{16}{p^2} \quad\Rightarrow\quad px + \frac{4}{p}x \:=\:p^2-\frac{16}{p^2}

    . . Factor: . \left(p + \frac{4}{p}\right)x \:=\:\left(p - \frac{4}{p}\right)\left(p + \frac{4}{p}\right) \quad\Rightarrow\quad x \:=\:p - \frac{4}{p}

    Substitute into [1]: . y \:=\:p\left(p - \frac{4}{p}\right) - p^2 \quad\Rightarrow\quad \boxed{y \:=\:-4}

    The locus of T is a horizontal line!




    c) If M is the midpoint of chord PQ, determine the locus of M.
    We have: . P(2p,p^2)\,\text{ and }\,Q\left(-\frac{8}{p},\:\frac{16}{p^2}\right)


    The midpoint has coordinates:

    . . x \;=\;\frac{2p-\frac{8}{p}}{2} \;=\;p - \frac{4}{p}\;\;{\color{blue}[3]}\qquad\qquad y \;=\;\frac{p^2 + \frac{16}{p^2}}{2} \;=\;\frac{p^4+16}{2p^2}\;\;{\color{blue}[4]}


    Square [3]: . x^2 \;\;=\;\;p^2 - 8 + \frac{16}{p^2} \;\;=\;\;\frac{p^4 + 16}{p^2} - 8 \;\;=\;\;2\left(\frac{p^4+16}{2p^2}\right) - 8

    Substitute [4]: . \boxed{x^2 \;=\;2y - 8}


    The locus of M is another parabola,
    . . somewhat "narrrower" with vertex (0,4).

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