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$\displaystyle R$ is the fixed point (3,2), and $\displaystyle P$ is the movable point (x,y)
Find the equation of the locus of $\displaystyle P$
if distance $\displaystyle PR$ is twice the distance from $\displaystyle P$ to $\displaystyle y=1$ Code:

 P
 o (x,y)
 * :
 * :
 * :
 R o :
 (3,2) :
+:
 :Q
 +         o  
 (x,1)

We have: .$\displaystyle PR \:=\:2\!\cdot\!PQ \quad\Rightarrow\quad \sqrt{(x3)^2+(y2)^2} \;=\;2(y+1)$
Square both sides: .$\displaystyle (x3)^2 + (y2)^2 \;=\;4(y1)^2$
Expand: .$\displaystyle x^26x + 9 + y^2  4y + 4 \;=\;4y^2 + 8y + 4$
. . . . . . .$\displaystyle x^2  6x  3y^3  12y \;=\;9$
Complete the square:
. . $\displaystyle (x^2  6x \qquad)  3(y^2 + 4y \qquad) \;=\;9$
. . $\displaystyle (x^2  6x \:{\color{blue}+\:9})  3(y^2 + 4y \:{\color{red}+\:4}) \;=\;9 \:{\color{blue}+\:9} \: {\color{red}\:12}$
. . . . . . . . $\displaystyle (x3)^2  3(y+2)^2 \;=\;12$
Divide by 12: .$\displaystyle \boxed{\;\frac{(y+2)^2}{4}  \frac{(x3)^2}{12} \;=\;1\;}$