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Math Help - The Parabola

  1. #1
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    The Parabola

    Currently stuck on my hw, to be handed in on Monday, wondering if anyone can help?

    The normal to the parabola with the equation y^2=4ax at the point P(at^2, 2at) meets the curve again at the point Q (as^2, 2as). Show that s=-t-(2/t).
    I'm not sure where the a, x and y have dissappeared to because when I worked out the equation to the normals at P and Q, I couldn't eliminate them.

    Can someone shed any light on this please?
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  2. #2
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    y^{2}=4ax

    P(at^{2}, \;\ 2at), \;\ Q(as^{2}, \;\ 2as)

    The slope of the tangent line is:

    2yy'=4a

    y'=\frac{4a}{2y}

    The slope at P is \frac{4a}{2(2at)}=\frac{1}{t}

    Therefore, the slope of the normal line is -t.

    The line equation is 2at=-t(at^{2})+b

    y=-tx+2at+at^{3}


    Use the coordinates of Q and solve for s:

    2as=-t(as^{2})+2at+at^{3}

    tas^{2}+2as = 2at+at^{3}

    Factor:

    a(s-t)(t^{2}+st+2)

    s=0 when s=t or upon solving the quadratic: t^{2}+st+2=0

    \boxed{s = -t-\frac{2}{t}}
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