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The parametric equations of a curve are x = 4t and y = 4 − t2.

Find the equations of the normals to the curve at the points where
the curve meets the x-axis. Hence, find the point of intersection of these normals.

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  1. I bet you, that second equation is y = 4-t^2

    form the first: x^2 = 16t^2 --> t^2 = x^2/16
    from the 2nd: t^2 = 4 - y

    so x^2/16 = 4-y
    x^2 = 64 - 16y , which is a parabola opening downwards.

    derivative: 2x = -16 dy/dx
    dy/dx = -x/8

    x-intercepts of the parabola, let y = 0
    x = ±8
    so we have intercepts at (8,0) and (-8,0)

    at (8,0), slope of tangent = -8/8 = -1
    so the normal has a slope of +1
    equation of normal: y = 1(x-8)

    at (-8,0), slope of tangent = 1
    so slope of normal = -1
    equation of normal : y = -(x+8)

    solving these two normals:
    x-8 = -x-8
    2x = 0
    x = 0 , obvious from the symmetry
    then y = -8

    the normals intersect at (0,-8)

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