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    Mathematics - parametric equations?
    A curve is defined by the parametric equations x = 2t^2, y = 4t
    (i)By eliminating the parameter, find the cartesian equation of the curve.
    (ii) Find the equation of the tangent to the curve at the point A with parameter t = 2
    (iii) show that the tangent does not meet the curve again.
    (iv) the normal of the curve at A cuts the curve again at B. Find the coordinates of B.



    In the standard spirit of TSR, I'll post what I've done so far so you know where to help me.

    i) x = 2t^2, y = 4t
    t=y/4
    x = (1/8)y^2
    y= sqrt(8x)

    ii) dy/dx = (dy/dt)/(dx/dt) = 4/4t. When t=2, dy/dx = 1/2.

    Using y=mx+c,

    y=0.5x+c

    Where do I go from here please? I'm confused with the extra t.
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    (Original post by jumblehunter)
    Mathematics - parametric equations?
    A curve is defined by the parametric equations x = 2t^2, y = 4t
    (i)By eliminating the parameter, find the cartesian equation of the curve.
    (ii) Find the equation of the tangent to the curve at the point A with parameter t = 2
    (iii) show that the tangent does not meet the curve again.
    (iv) the normal of the curve at A cuts the curve again at B. Find the coordinates of B.



    In the standard spirit of TSR, I'll post what I've done so far so you know where to help me.

    i) x = 2t^2, y = 4t
    t=y/4
    x = (1/8)y^2
    y= sqrt(8x)

    ii) dy/dx = (dy/dt)/(dx/dt) = 4/4t. When t=2, dy/dx = 1/2.

    Using y=mx+c,

    y=0.5x+c

    Where do I go from here please? I'm confused with the extra t.

    Put t=2 into x and y to find your point
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    what values do you get for x and y at the value t=2?
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    (Original post by TenOfThem)
    Put t=2 into x and y to find your point
    so 8=4+c
    c=4?
    y=0.5x+4?


    You sure thats right. Thats not what I've been told the answer is. http://uk.answers.yahoo.com/question...0113754AAcHGIu
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    (Original post by jumblehunter)
    so 8=4+c
    c=4?
    y=0.5x+4?


    You sure thats right. Thats not what I've been told the answer is. http://uk.answers.yahoo.com/question...0113754AAcHGIu
    That answer is the same
 
 
 
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