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Math Help - locus of the mid-point

  1. #1
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    locus of the mid-point

    question: Avariable line passing through the point (5,0) intersects the lines 3x-4y=0 and 3x+4y=0 at H and K respectively. Find the equaion of the locus of the mid point of HK.

    here is my solution:
    let P (x,y) be the mid-point
    y/(x-5)=m , where m is the slope of variable line
    y = mx-5m
    put y = mx-5m to 3x-4y=0,
    3x-4(mx-5m)=0
    x=20m/(4m-3)
    put y = mx-5m to 3x+4y=0,
    3x+4(mx-5m)=0
    x=20m/(4m+3)
    since P is the mid-point of HK
    x-coordinate of P = 0.5[20m/(4m-3)+ 20m/(4m+3)]
    =(80m^2)/(16m^2-9)
    put m=y/(x-5)
    x=80[y-(x-5)^2)]/[16(y/x-5)^2-9]
    90x^3-90x^2+80y^2 -16xy^2+225x=0 is my equation of the locus
    but the answer is my textbook says it is 16y^2=9x(x-5)

    here is the solution of my textbook:
    let H be (4s,3s), K be (4t,-3t).
    Slope of HK = (3s+3t)/(4s-4t) = [3(3+t)]/[4(s-t)]
    it also equals 3s/(4s-5)
    therefore: [3(3+t)]/[4(s-t)]=3s/(4s-5)
    8st=5(s+t)
    let P(x,y) be the mid -point of HK,
    x=(4s+4t)/2=2(s+t) ------------------ (1)
    y=3(s-t)/2 -------------------------- (2)
    eliminating s and t from (1)and(2):
    (3x)^2-(4y)^2 = [6(s+t)]^2 - [6(s-t)]^2
    16y^2 = 9x(x-5)

    i cannot figure out how can it get (3x)^2-(4y)^2 = [6(s+t)]^2 - [6(s-t)]^2 in order to do the elimination of s and t (seems it could do it just because it knows the final ans.) it also use a different method from mine.
    furthermore i want to know my solution is wrong, can anyone point it out?
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  3. #3
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    Quote Originally Posted by afeasfaerw23231233 View Post
    [Your answer (correcting you typo)]
    9x^3-90x^2+80y^2 -16xy^2+225x=0 is my equation of

    [Book's answer]
    16y^2 = 9x(x-5)
    Note that
    9x^3-90x^2+80y^2 -16xy^2+225x = (x - 5)(9x^2 - 45x - 16y^2)

    So unless x = 5 your answer and the book's answer are the same.

    Now you tell me why your solution method fails to produce an answer for x = 5.

    -Dan
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    the question is to ask the equation of the locus of the mid-point of HK.
    so this means my answer 9x^3-90x^2+80y^2 -16xy^2+225x=0 is the same as my book's 16y^2 = 9x(x-5)?
    and i also want to know if my method if correct, though it seems tedious and slow while compared to the one in my book's answer.
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  5. #5
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by afeasfaerw23231233 View Post
    the question is to ask the equation of the locus of the mid-point of HK.
    so this means my answer 9x^3-90x^2+80y^2 -16xy^2+225x=0 is the same as my book's 16y^2 = 9x(x-5)?
    and i also want to know if my method if correct, though it seems tedious and slow while compared to the one in my book's answer.
    The only thing I see wrong with your method (and I would have done it the same way you did) is that it doesn't generalize to the case where the variable line is vertical. This is the origin of the x \neq 5 restriction in your solution compared your book's answer.

    As to why your equation is so much more complicated I can't really say.

    -Dan
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