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Math Help - Equation of Locus

  1. #1
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    Equation of Locus

    I am not able to solve this problem-

    Two fixed points A and B are taken on the axes such that
    OA = a and OB = h; two variable points A' and B' are taken on the
    same axes; find the locus of the intersection of AB' and A'B
    (1) when OA' + OB' = OA + OB,
    and (2) when 1/OA' - 1/OB' = 1/OA - 1/OB


    What I was doing
    OA' + OB' = a + b
    a- OA' + b - OB' = 0
    AA' + BB' = 0
    AA'=-BB'

    OA' = OA -AA'
    = a + BB'
    x/(a + BB') +y/b = 1 equation No. 1

    OB' = OB - BB'
    = b - BB'
    x/(a) +y/(b-BB') = 1 equation No. 2

    Solving 1 and 2 to eliminate BB' and get a relation of required locus.

    But answer is of (i) x+y-a-b=0
    (ii) y=x
    Thanx in Advance.
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  2. #2
    MHF Contributor
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    Re: Equation of Locus

    Quote Originally Posted by varunkanpur View Post
    I am not able to solve this problem-

    Two fixed points A and B are taken on the axes such that
    OA = a and OB = h; two variable points A' and B' are taken on the
    same axes; find the locus of the intersection of AB' and A'B
    (1) when OA' + OB' = OA + OB,
    and (2) when 1/OA' - 1/OB' = 1/OA - 1/OB


    What I was doing
    OA' + OB' = a + b
    a- OA' + b - OB' = 0
    AA' + BB' = 0
    AA'=-BB'

    OA' = OA -AA'
    = a + BB'
    x/(a + BB') +y/b = 1 equation No. 1

    OB' = OB - BB'
    = b - BB'
    x/(a) +y/(b-BB') = 1 equation No. 2

    Solving 1 and 2 to eliminate BB' and get a relation of required locus.

    But answer is of (i) x+y-a-b=0
    (ii) y=x
    Thanx in Advance.
    First off I assume you are talking about line segments BA' and B'A, not the entire line defined by those points.

    What I did to solve (a) was to derive the formula for the lines B'A and BA' given an A'. Given that A' you know that B'=(a+h-a')
    What you find is that the segments only intersect when A'=A=a in which case the two lines are identical and the locus of intersection is the entire segment.
    This needs to be proven of course which would be done by finding the point of intersection and showing that it lies outside the segment endpoints.

    I haven't looked at (b) in detail yet but the same method should apply.
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  3. #3
    Super Member
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    Athens, OH, USA
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    Re: Equation of Locus

    Hi,
    Your question has different possible interpretations. Romsek has given one possibility. Another interpretation is that the lines A'B and AB' are infinite lines which intersect in exactly one point (so different non-parallel lines). The attachment shows a solution for i) under this interpretation (details left to you):

    Equation of Locus-mhfgeometry58.png
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