Results 1 to 2 of 2

Thread: Parametric Equations

  1. #1
    Junior Member
    Joined
    Apr 2008
    Posts
    45

    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 $\displaystyle x^2 = 4y$, 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.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Super Member

    Joined
    May 2006
    From
    Lexington, MA (USA)
    Posts
    12,028
    Thanks
    848
    Hello, Flay!

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

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

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

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

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

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


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




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


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

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

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

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

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

    The locus of $\displaystyle T$ is a horizontal line!




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


    The midpoint has coordinates:

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


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

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


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

    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. parametric equations, equations of plane
    Posted in the Calculus Forum
    Replies: 1
    Last Post: Sep 10th 2009, 02:58 AM
  2. Parametric equations to rectangular equations.
    Posted in the Pre-Calculus Forum
    Replies: 2
    Last Post: Apr 5th 2009, 10:39 PM
  3. Parametric equations
    Posted in the Calculus Forum
    Replies: 1
    Last Post: Mar 15th 2009, 02:18 PM
  4. Replies: 3
    Last Post: Dec 2nd 2008, 10:54 AM
  5. Replies: 1
    Last Post: Sep 1st 2007, 06:35 AM

Search Tags


/mathhelpforum @mathhelpforum