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Math Help - Another question about the parabola...

  1. #1
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    Another question about the parabola...

    Lol I seem to be stuck with a lot of questions to do with the parabola.

    Q: Show that the normals at P (at^2, 2at) and Q (as^2,2as)on the parabola with equation y^2=4ax meet at the point R where the coordinates of R are
    [a(t^2+ts+s^2+2)], -ats(t+s)]. This part of the question I've managed to do. It's the next part that I'm stuck on: Given that the line PQ passes through the focus S(a, 0), show that as t and s vary, R lies on the curve with equation y^2=a(x-3a).

    I tried to eliminate the variables t and s, but was unsuccessful. Can someone help please?

    Btw, can someone let me know what topic this post should come under for future reference?
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  2. #2
    MHF Contributor red_dog's Avatar
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    The line PQ has the equation
    \displaystyle\frac{x-at^2}{a(s^2-t^2}=\frac{y-2at}{2a(s-t)}
    or \displaystyle\frac{x-at^2}{s+t}=\frac{y-2at}{2}.
    If the line passes through the focus, then we have
    \displaystyle\frac{a-at^2}{s+t}=-\frac{2at}{2}\Rightarrow st=-1.
    The coordinates of R are
    x=a(t^2+st+s^2+2)\Rightarrow x=a(t^2+s^2+1)
    y=-ats(t+s)\Rightarrow y=a(t+s).
    Now, we have
    y^2=a^2(t^2+s^2+2st)=a^2(t^2+s^2+1-3)=a^2\left(\frac{x}{a}-3\right)=a(x-3a)
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