Q&A
Your initial formula was correct, however, you made a few mistakes in the working.
$\displaystyle 4 PA^2 = PB $
$\displaystyle 4 [ (y-1)^2 + (x-0)^2] = (y+2)^2 + (x-0)^2 $
$\displaystyle 4y^2 - 8y + 4 + 4x^2$ (you forgot to multiply 4 to x²) = $\displaystyle y^2 + 4y + x^2 $
$\displaystyle 4y^2 - y^2 -8y - 4y$ (you added the 4y, instead of subtracting) $\displaystyle + 4 -4 + 4x^2 - x^2 = 0 $
$\displaystyle 3y^2 + 3x^2 -12y = 0 $
Divide everything by 3 to get the answer required
Part B is relatively straight forward. Substitute point Q into the equation of the locus and check if equality holds (it should)
Then simply find the gradient between two points you already know (B and Q) and then use the point gradient formula.
Hello, nikk!
You didn't "carry the 4."Given: points $\displaystyle A(0,1)\text{ and }B(0,-2).$
Point $\displaystyle P(x,y)$ moves such that the ratio of .$\displaystyle PA:PB \,=\,1:2$
(a) Show that the equation of locus $\displaystyle P$ is: .$\displaystyle x^2+y^2-4y \:=\:0$
$\displaystyle 2\cdot PA \:=\:PB \quad\Rightarrow\quad 2\sqrt{(y-1)^2 + x^2} \;=\;\sqrt{(y+2)^2+x^2}$
. . $\displaystyle 4{\color{red}(}y^2-2y+1+x^2{\color{red})} \;=\;y^2+4y + 4 + x^2$
. . .$\displaystyle 4y^2 - 8y + 4 + 4x^2 \;=\;y^2 + 4y + 4 + x^2$
. . . . $\displaystyle 3x^2 + 3y^2 - 12y \;=\;0$
. . . . . . $\displaystyle x^2 + y^2 - 4y \;=\;0$
Substitute $\displaystyle (2,2)$ into: $\displaystyle x^2+y^2-4y \:=\:0$(b) Show that point $\displaystyle Q(2,2)$ is a point of the locus $\displaystyle P.$
. . $\displaystyle 2^2 + 2^2 - 4(2) \;=\;0 \quad\hdots \text{ True!}$
The slope of $\displaystyle BQ$ is: .$\displaystyle m \:=\:\frac{2-(-2)}{2-0} \:=\:2$(c) Find the equation of straiggt line $\displaystyle BQ.$
The equation of the line through $\displaystyle B(0,\text{-}2)$ with slope $\displaystyle m = 2$ is:
. . $\displaystyle y - (\text{-}2) \:=\:2(x-0) \quad\Rightarrow\quad y \:=\:2x-4$