Let [tex](x,y)[/tex] be the coordinates of point P.
The distance from the x-axis is exactly the y coordinates of point P.
The distance from the point (3,-3) is given by
[tex]\sqrt{(x-3)^2+(x+3)^2}[/tex]
So, we want
[tex]y = 3\sqrt{(x-3)^2+(y+3)^2}[/tex]
We can square both sides to get
[tex]y^2=9((x-3)^2+(y+3)^2)[/tex]
Expanding the squares, we have
[tex]y^2=9(x^2-6x+9+y^2+6y+9) = 9(x^2+y^2-6x+6y+18)=9x^2+9y^2-54x+54y+162[/tex]
So, the final equation is
[tex]y^2=9x^2+9y^2-54x+54y+162 \iff 9x^2+8y^2-54x+54y+162=0[/tex]
