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    The curve C has parametric equations

    x=4cos2t, y=3sint, -90 < t < 90

    A is the point(2, 1 1/2) and lies on C.

    a) Find the value of t at the point A

    b) find dy/dx in terms of t.

    c) Show that an equation of the normal to C at A is 6y - 16x + 23 = 0

    The normal at A cuts C again at the point B.

    d) Find the y-coordinate of the point B

    Done a) (answer = 30) b) (answer = 3cost/-8sin2t) and c) but Im stuck on d)

    Can anyone help? How do I get a cartesian equation from the curve C...

    stick your parametric values of x and y into the equation of the normal, you should get two values of t, one for the point A and one for the point B, and use that to find the y-coordinate for B
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    (Original post by imasillynarb)
    The curve C has parametric equations

    x=4cos2t, y=3sint, -90 < t < 90

    A is the point(2, 1 1/2) and lies on C.

    a) Find the value of t at the point A

    b) find dy/dx in terms of t.

    c) Show that an equation of the normal to C at A is 6y - 16x + 23 = 0

    The normal at A cuts C again at the point B.

    d) Find the y-coordinate of the point B

    Done a) (answer = 30) b) (answer = 3cost/-8sin2t) and c) but Im stuck on d)

    Can anyone help? How do I get a cartesian equation from the curve C...
    a) 2=4cos2t
    1/2=cos2t
    arccos 1/2 = 2t = 60 => t = 30

    b) dy/dt = 3cost
    dx/dt = -8sin2t
    dy/dx = -3cost/8sin2t

    c) dy/dx at A = -3cos30/8sin60 = -3rt3/2/8rt3/2 = -3/8
    m for the normal = 8/3 so y-1.5 = 8/3(x-2) = 8x/3-16/3
    => 6y-9 = 16x-32 => 6y-16x+23 = 0

    Sorry, didn't notice you did these, oops...

    And I'm also stuck on d. But this may be useful...
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    (Original post by GeneralGrievous)
    stick your parametric values of x and y into the equation of the normal, you should get two values of t, one for the point A and one for the point B, and use that to find the y-coordinate for B
    I didnt quite understand that, can you go into abit more detail?

    6(3sint) - 16(4cos2t) + 23 = 0

    => 18 sint -64(1-2sin²t) + 23 = 0

    => 128 sin²t + 18 sint - 41 = 0

    sint = (-18 +- sqrt(324 + 4.128.41))/2.128

    sint = -0.64, 0.5

    [0.5 corresponds to t=30, ie point A]

    y = 3 sint = 3*-0.64 = -1.92
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    Nice one, cheers
 
 
 
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