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Math Help - Differentiation

  1. #1
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    Differentiation

    The curve C has parametric equations

    x = 4 cos 2t, y = 3 sin t, -pi/2 < t < pi/2

    A is the point (2, 3/2), and lies on C.

    a) 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.
    b) Find the y-coordinate of the point B.

    Ive done a, but how can I find b. Please help me.
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  2. #2
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    Quote Originally Posted by geton View Post
    The curve C has parametric equations

    x = 4 cos 2t, y = 3 sin t, -pi/2 < t < pi/2

    A is the point (2, 3/2), and lies on C.

    a) 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.
    b) Find the y-coordinate of the point B.
    To do the second part if the parametrized curve intersects the line then x=4\cos 2t and y=3\sin t for t\in (-\pi/2,\pi/2) which means 18\sin t - 64 \cos 2t +23 = 0. Solve for t on this interval.
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  3. #3
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    Quote Originally Posted by ThePerfectHacker View Post
    18\sin t - 64 \cos 2t +23 = 0. Solve for t on this interval.
    I must say Im stupid. Thank you for this line. Problem solved
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