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Thread: equidistant problem

  1. #1
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    equidistant problem

    Question: A point P(a,b) is equidistant from the y-axis and from the point (4,0). find a relationship between a and b

    My attempt
    using the formula to find the coordinates on y-axis
     m=[(\frac{x_{1}+x_{2}}{2}),(\frac{y_{1}+y_{2}}{2})]

    find the a
    since the coordinate is on the y-axis therefore x= 0
    hence
     \frac{0+4}{2} =2 = a therefore a = 2

    since (4,0) P(a,b)
     \frac{0 + y_{2}}{2} =b

    therefore y=2b
    the coordinates on the y-axis is (0,2b) let 's call it b and a= (4,0)

    find the distance of BP and AP
     BP = \sqrt{(2b-b)^2 +(0-a)^2}
    BP = \sqrt{b^2+a^2}

    AP = \sqrt{(4-a)^2 +(0-b)^2}
    AP = \sqrt{16-8a+a^2 +b^2}


    find AB
     \sqrt{(4-0)^2+(0-2b)^2} \sqrt{16+4b^2}

    AB = AP + BP
    16+4b^2 = 16-8a+a^2 +b^2 +b^2+a^2
    b^2 = a^2-4a

    the answer in the book is
    b^2 = 8a -16
    where did i go wrong?
    please any help
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  2. #2
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    Re: equidistant problem

    let's used squared distance instead of distance for convenience

    the squared distance of $(a,b)$ to the y-axis is $a^2$

    the squared distance of $(a,b)$ to the point $(4,0)$ is $(a-4)^2 + b^2$

    set these two equal

    $(a-4)^2 + b^2 = a^2$

    $a^2 -8a + 16 + b^2 = a^2$

    $b^2 = 8a - 16$
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  3. #3
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    Re: equidistant problem

    Quote Originally Posted by romsek View Post
    let's used squared distance instead of distance for convenience

    the squared distance of $(a,b)$ to the y-axis is $a^2$

    the squared distance of $(a,b)$ to the point $(4,0)$ is $(a-4)^2 + b^2$

    set these two equal

    $(a-4)^2 + b^2 = a^2$

    $a^2 -8a + 16 + b^2 = a^2$

    $b^2 = 8a - 16$
    I thought this question as this equidistant problem-yh.jpg but after your correction for I noticed that the question had this sketch in mind equidistant problem-yh1.jpg
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  4. #4
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    Re: equidistant problem

    Quote Originally Posted by romsek View Post
    let's used squared distance instead of distance for convenience

    the squared distance of $(a,b)$ to the y-axis is $a^2$

    the squared distance of $(a,b)$ to the point $(4,0)$ is $(a-4)^2 + b^2$

    set these two equal

    $(a-4)^2 + b^2 = a^2$

    $a^2 -8a + 16 + b^2 = a^2$

    $b^2 = 8a - 16$
    thnks
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