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Math Help - Equation Of A Locus

  1. #1
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    Equation Of A Locus

    ,
    I'm having trouble dealing with the following question. I need your help.

    Q) Find the equation of the locus of a point P whose distance from (-1,1) is equal to thrice it's distance from the Y-axis.

    The answer is reported as 8x^2 - y^2 -2x +2y -2 = 0, which i failed to get.
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  2. #2
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    Equation of a locus

    Hello saberteeth
    Quote Originally Posted by saberteeth View Post
    ,
    I'm having trouble dealing with the following question. I need your help.

    Q) Find the equation of the locus of a point P whose distance from (-1,1) is equal to thrice it's distance from the Y-axis.

    The answer is reported as 8x^2 - y^2 -2x +2y -2 = 0, which i failed to get.
    Suppose the point P is (x,y). Then its distance from (-1,1) is \sqrt{(x+1)^2+(y-1)^2}, and its distance from the y-axis is simply x. So its locus has equation:

    \sqrt{(x+1)^2+(y-1)^2} = 3x

    i.e. (x+1)^2 +(y-1)^2 = 9x^2 (Did you forget to square the 3?)

    i.e. x^2+2x+1+y^2-2y+1 = 9x^2

    i.e. 8x^2 -y^2 -2x+2y -2 =0

    Grandad
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  3. #3
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    Quote Originally Posted by Grandad View Post
    Hello saberteethSuppose the point P is (x,y). Then its distance from (-1,1) is \sqrt{(x+1)^2+(y-1)^2}, and its distance from the y-axis is simply x. So its locus has equation:

    \sqrt{(x+1)^2+(y-1)^2} = 3x

    i.e. (x+1)^2 +(y-1)^2 = 9x^2 (Did you forget to square the 3?)

    i.e. x^2+2x+1+y^2-2y+1 = 9x^2

    i.e. 8x^2 -y^2 -2x+2y -2 =0

    Grandad
    Thanks Grandad. I was on the right track but i goofed up somewhere which i need to check.
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