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Math Help - Locus question

  1. #1
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    Locus question

    Hi,
    I have this problem:

    If 'R' is the fixed point (3,2) and 'P' is the movable point (x,y)
    Find the equation of the locus of 'P' if distance PR is twice the distance from P to y=-1

    Thanks!
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  2. #2
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    Quote Originally Posted by withyou View Post
    Hi,
    I have this problem:

    If 'R' is the fixed point (3,2) and 'P' is the movable point (x,y)
    Find the equation of the locus of 'P' if distance PR is twice the distance from P to y=-1

    Thanks!
    PR=\sqrt{(x-3)^2 + (y-2)^2}.

    Distance of P from the line y = -1 is y + 1.

    Therefore ......
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  3. #3
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    Hello, withyou!

    Did you make a sketch?
    Exactly where is your difficulty?


    R is the fixed point (3,2), and P is the movable point (x,y)

    Find the equation of the locus of P
    if distance PR is twice the distance from P to y=-1
    Code:
            |
            |                 P
            |                 o (x,y)
            |              *  :
            |           *     :
            |        *        :
            |   R o           :
            |   (3,2)         :
        ----+-----------------:------
            |                 :Q
          - + - - - - - - - - o - -
            |               (x,-1)
            |

    We have: . PR \:=\:2\!\cdot\!PQ \quad\Rightarrow\quad \sqrt{(x-3)^2+(y-2)^2} \;=\;2(y+1)

    Square both sides: . (x-3)^2 + (y-2)^2 \;=\;4(y-1)^2

    Expand: . x^2-6x + 9 + y^2 - 4y + 4 \;=\;4y^2 + 8y + 4

    . . . . . . . x^2 - 6x - 3y^3 - 12y \;=\;-9


    Complete the square:

    . . (x^2 - 6x  \qquad) - 3(y^2 + 4y \qquad) \;=\;-9

    . . (x^2 - 6x \:{\color{blue}+\:9}) - 3(y^2 + 4y \:{\color{red}+\:4}) \;=\;-9 \:{\color{blue}+\:9} \: {\color{red}-\:12}

    . . . . . . . . (x-3)^2 - 3(y+2)^2 \;=\;-12


    Divide by -12: . \boxed{\;\frac{(y+2)^2}{4} - \frac{(x-3)^2}{12} \;=\;1\;}

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