1. ## locus, need confirmation

Q&A

2. Your initial formula was correct, however, you made a few mistakes in the working.

$4 PA^2 = PB$

$4 [ (y-1)^2 + (x-0)^2] = (y+2)^2 + (x-0)^2$

$4y^2 - 8y + 4 + 4x^2$ (you forgot to multiply 4 to x²) = $y^2 + 4y + x^2$

$4y^2 - y^2 -8y - 4y$ (you added the 4y, instead of subtracting) $+ 4 -4 + 4x^2 - x^2 = 0$

$3y^2 + 3x^2 -12y = 0$

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.

3. Hello, nikk!

Given: points $A(0,1)\text{ and }B(0,-2).$
Point $P(x,y)$ moves such that the ratio of . $PA:PB \,=\,1:2$

(a) Show that the equation of locus $P$ is: . $x^2+y^2-4y \:=\:0$
You didn't "carry the 4."

$2\cdot PA \:=\:PB \quad\Rightarrow\quad 2\sqrt{(y-1)^2 + x^2} \;=\;\sqrt{(y+2)^2+x^2}$

. . $4{\color{red}(}y^2-2y+1+x^2{\color{red})} \;=\;y^2+4y + 4 + x^2$

. . . $4y^2 - 8y + 4 + 4x^2 \;=\;y^2 + 4y + 4 + x^2$

. . . . $3x^2 + 3y^2 - 12y \;=\;0$

. . . . . . $x^2 + y^2 - 4y \;=\;0$

(b) Show that point $Q(2,2)$ is a point of the locus $P.$
Substitute $(2,2)$ into: $x^2+y^2-4y \:=\:0$

. . $2^2 + 2^2 - 4(2) \;=\;0 \quad\hdots \text{ True!}$

(c) Find the equation of straiggt line $BQ.$
The slope of $BQ$ is: . $m \:=\:\frac{2-(-2)}{2-0} \:=\:2$

The equation of the line through $B(0,\text{-}2)$ with slope $m = 2$ is:

. . $y - (\text{-}2) \:=\:2(x-0) \quad\Rightarrow\quad y \:=\:2x-4$

4. ## Tq

Gusbob & Soroban tq for your help and kindness