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Math Help - 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 x = at^2 , y = 2at. Show that the equation of the normal to C at the point P, whose parameter is p, is:

    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:

    dx/dt = 2at
    dy/dt = 2a

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

    Normal Gradient = -1/m = -p

    (Use p as parameter for normal equation)

    y - y1 = m(x - x1)
    y - 2ap = -p(x - ap^2)
    y - 2ap = -px + ap^3
    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:

    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 x = at^2 , y = 2at. Show that the equation of the normal to C at the point P, whose parameter is p, is:

    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:



    dx/dt = 2at


    dy/dt = 2a



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



    Normal Gradient = -1/m = -p



    (Use p as parameter for normal equation)



    y - y1 = m(x - x1)


    y - 2ap = -p(x - ap^2)


    y - 2ap = -px + ap^3


    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:



    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 y=0 in the line you get 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 (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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