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Thread: Finding Intersection of Parametric Equation and Axis

  1. #1
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    Finding Intersection of Parametric Equation and Axis

    Hi, I've been trying to learn the C4 paper ahead of schedule and have gotten down to one last question I can't quite get.

    Q5) A curve C has parametric equations $\displaystyle x = at^2$ , $\displaystyle y = 2at$. Show that the equation of the normal to C at the point P, whose parameter is $\displaystyle p$, is:

    $\displaystyle px + y - 2ap - ap^3 = 0$

    The normal to C at P meets the x-axis at Q. The perpendicular from P to the x-axis meets the x-axis at R. Find the length of QR.

    This is my working so far:

    $\displaystyle dx/dt = 2at$
    $\displaystyle dy/dt = 2a$

    $\displaystyle dy/dx = dy/dt * dt/dx = 1/t$ (Chain Rule)

    Normal Gradient = $\displaystyle -1/m$ = $\displaystyle -p$

    (Use p as parameter for normal equation)

    $\displaystyle y - y1 = m(x - x1)$
    $\displaystyle y - 2ap = -p(x - ap^2)$
    $\displaystyle y - 2ap = -px + ap^3$
    $\displaystyle px + y - 2ap - ap^3 = 0$ (As required)

    And thats about as far as I got, I've been trying to let y = 0 and let q be the parameter, so:

    $\displaystyle qx - 2aq - aq^3 = 0$

    But do you then make them equal to each other and figure it out from there? I don't really know, its a WJEC paper too so I can't get any marking schemes without paying, so I thought this way would be better to get an understanding.

    Thanks




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  2. #2
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    Quote Originally Posted by steve989 View Post
    Hi, I've been trying to learn the C4 paper ahead of schedule and have gotten down to one last question I can't quite get.

    Q5) A curve C has parametric equations $\displaystyle x = at^2$ , $\displaystyle y = 2at$. Show that the equation of the normal to C at the point P, whose parameter is $\displaystyle p$, is:

    $\displaystyle px + y - 2ap - ap^3 = 0$


    The normal to C at P meets the x-axis at Q. The perpendicular from P to the x-axis meets the x-axis at R. Find the length of QR.



    This is my working so far:



    $\displaystyle dx/dt = 2at$


    $\displaystyle dy/dt = 2a$



    $\displaystyle dy/dx = dy/dt * dt/dx = 1/t$ (Chain Rule)



    Normal Gradient = $\displaystyle -1/m$ = $\displaystyle -p$



    (Use p as parameter for normal equation)



    $\displaystyle y - y1 = m(x - x1)$


    $\displaystyle y - 2ap = -p(x - ap^2)$


    $\displaystyle y - 2ap = -px + ap^3$


    $\displaystyle px + y - 2ap - ap^3 = 0$ (As required)



    And thats about as far as I got, I've been trying to let y = 0 and let q be the parameter, so:



    $\displaystyle qx - 2aq - aq^3 = 0$



    But do you then make them equal to each other and figure it out from there? I don't really know, its a WJEC paper too so I can't get any marking schemes without paying, so I thought this way would be better to get an understanding.




    Thanks

    So you do the hard part and get stuck in an almost trivial part? Putting $\displaystyle y=0$ in the line you get $\displaystyle x=2a+ap^2\Longrightarrow Q=(2a+ap^2,0)$ , and the perpendicular from the point P to the x-axis meets the axis at $\displaystyle (ap^2,0)$ , so...

    Tonio
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  3. #3
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    Oh yeah, haha that put me in my place, thanks a lot, was a long day, spent 4-5 hours researching unknown topics

    Thanks again
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