Locus question

**withyou** · July 19th 2008, 03:10 AM

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!

**mr fantastic** · July 19th 2008, 05:17 AM

Originally Posted by withyou

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 ......

**Soroban** · July 19th 2008, 05:30 AM

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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