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    A curve is given by x = 2t + 3, y = t^3 - 4t, where t is a parameter. The point A has parameter t = -1 and the line l is the tangent to C at A. The line L also cuts the curve at B.
    a. Show that an equation for L is 2y + x = 7
    b. Find the value of t at B


    Part a is fine but part b?
    I tried converting the parametric equations into a cartesian equation then doing simultaneous equations but I know that's wrong?

    Help?
    Thanks
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    you have a set of 3 simultaneous equations, and 3 unknowns.
    x=2t+3
    y=t^3-4t
    2y+x=7

    Solving them is slightly more involved than solving a pair of simultaneous equations with only 2 unknowns but it's very possible, and it's not wrong, point B satisfies all of these equations.
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    You can substitute x = 2t + 3 and y = t^3 - 4t into 2y + x = 7, form a cubic equation, there you will have to use long division to factorise the cubic.
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    if you differentiate to find dy/dx you get (3t^2 - 4)*(-1/2) = 1.5t^2 - 2

    sub in the gradient you got from (a) = -1/2 and you get 1.5t^2 -2=-0.5
    so 1.5t^2 = 1.5
    t^2 = 1
    t = 1, -1
    so B is t=1
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    (Original post by titsmcgee)
    if you differentiate to find dy/dx you get (3t^2 - 4)*(-1/2) = 1.5t^2 - 2

    sub in the gradient you got from (a) = -1/2 and you get 1.5t^2 -2=-0.5
    so 1.5t^2 = 1.5
    t^2 = 1
    t = 1, -1
    so B is t=1
    You cannot take the gradient at B to be -0.5, as the line CUTS at B, not touches so it is not a tangent at B.
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    I've got the answer now By using the 3 simulatenous equations and long division.
    Thanks!
 
 
 
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Updated: February 12, 2010
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