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Math Help - Parabola help!

  1. #1
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    Parabola help!

    P(ap^2,2ap) Q(aq^2,2aq) are on the parabola y^2=4ax.
    If the chord PQ passes through the focus of the parabola and the normals at P and Q meet at R, show that the locus of R is given by y^2+3a^2=ax.

    Since the chord is a focal chord, pq=-1. and q=-1/p
    then Q(\frac{a}{p^2}, -\frac{2a}{p}).
    the equations of the normals are
    y-2ap=-p(x-ap^2)_1 and y+\frac{2a}{p}=\frac{1}{p}(x-\frac{a}{p^2})_2
    1 \rightarrow y=-px+ap^3+2ap and
    2 \rightarrow  y=\frac{x}{p}-\frac{a}{p^3}-\frac{2a}{p}
    solving these two
    \frac{x}{p}-\frac{a}{p^3}-\frac{2a}{p}=-px+ap^3+2ap
    x=a(t^2+1+\frac{1}{t^2})
    and y=a(t-\frac{1}{t})
    now my problem is I have the parametric equation for the locus of R but i don't know how to get the cartesian equation from here.
    Thanks
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  2. #2
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    Hello arze
    Quote Originally Posted by arze View Post
    P(ap^2,2ap) Q(aq^2,2aq) are on the parabola y^2=4ax.
    If the chord PQ passes through the focus of the parabola and the normals at P and Q meet at R, show that the locus of R is given by y^2+3a^2=ax.

    Since the chord is a focal chord, pq=-1. and q=-1/p
    then Q(\frac{a}{p^2}, -\frac{2a}{p}).
    the equations of the normals are
    y-2ap=-p(x-ap^2)_1 and y+\frac{2a}{p}=\frac{1}{p}(x-\frac{a}{p^2})_2
    1 \rightarrow y=-px+ap^3+2ap and
    2 \rightarrow  y=\frac{x}{p}-\frac{a}{p^3}-\frac{2a}{p}
    solving these two
    \frac{x}{p}-\frac{a}{p^3}-\frac{2a}{p}=-px+ap^3+2ap
    x=a(t^2+1+\frac{1}{t^2})
    and y=a(t-\frac{1}{t})
    now my problem is I have the parametric equation for the locus of R but i don't know how to get the cartesian equation from here.
    Thanks
    You've done all the hard work! All you need to do now is to say:
    y=a\left(t-\frac{1}{t}\right)

    \Rightarrow y^2 = a^2\left(t^2-2+\frac{1}{t^2}\right)
    =a^2\left(t^2+1+\frac{1}{t^2}-3\right)

    =a^2\left(\frac{x}{a}-3\right)


    =a^2x-3a^2

    Grandad
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  3. #3
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    no wonder! i knew i was missing something
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