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  1. #1
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    locus - parabola

    p127 q8
    Find the equation of the locus of the mid-point of a vairable chord thro' the point (a , 2a ) of the parabola y^2=4ax
    thanks


    edit: my working: let p be a point on y^2=4ax , P = (at^2 , 2at).
    let A = (a, 2a)
    equ of PA: \frac {y-2a}{x-a} = \frac{2at-2a}{at^2-a}
    y= \frac {2(x-a)}{t+1} +2a
    have no idea how to do
    Last edited by afeasfaerw23231233; June 2nd 2008 at 07:37 PM.
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  2. #2
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    Hello,

    Quote Originally Posted by afeasfaerw23231233 View Post
    p127 q8
    Find the equation of the locus of the mid-point of a vairable chord thro' the point (a , 2a ) of the parabola y^2=4ax
    thanks


    edit: my working: let p be a point on y^2=4ax , P = (at^2 , 2at).
    let A = (a, 2a)
    equ of PA: \frac {y-2a}{x-a} = \frac{2at-2a}{at^2-a}
    y= \frac {2(x-a)}{t+1} +2a
    have no idea how to do
    Hmmm you're both near to it and not

    Let P like you did, and A(a, 2a).

    The equation of the line is y=mx+n, but actually, you don't need it (if I read correctly).

    The midpoint has its coordinates equal to half the sum of the other coordinates. Let M be the midpoint.


    M \left(\frac{a+at^2}{2} ~,~\frac{2a+2at}{2}\right) ...

    Am I right ??
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  3. #3
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    thanks! what a fool i am..
    i misread the question. i thought it asked me to find the mid point of the intersection of the chord and the parabola. can i find the locus of the mid point if it is the case?
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  4. #4
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    Quote Originally Posted by afeasfaerw23231233 View Post
    thanks! what a fool i am..
    i misread the question. i thought it asked me to find the mid point of the intersection of the chord and the parabola. can i find the locus of the mid point if it is the case?
    I've drawn the locus of the midpoints of all chords with A as one endpoint.

    If the parabola has the equation

    y^2 = 4ax

    The focus has the coordinates F(a, 0) and the point A(a, 2a) is placed on the parabola. (In my example I've used a = 3)

    M_1 and M_2 are midpoints of the chords.

    The locus is a parabola too.
    Attached Thumbnails Attached Thumbnails locus - parabola-parab_midptlocus.gif  
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  5. #5
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    thanks. i didn't see that a,2a is always a point on the parabola. i thought it was a point not lying on it. next time i have to sub the value of the point to the equation of the parabola to see if it is a point on it first.
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