Maths problem: how to do, (a), (b), thanks?

(a) Let P=(x,y)
Then, F=(x,3)
=> FP=3-y
AP²=(x-2)² + y²
BP²=x² + (y- -2)² = x² + (y+2)²

If AP²+BP²=2FP², we have
(x-2)² + y² + x² + (y+2)² = 2(3-y)² 
2x²+2y²-4x+4y+8 = 2(9-6y+y²)
x²+y²-2x+2y+4 = 9-6y+y²
x²-2x+8y-5 = 0

∴ eqt. of locus of P : x²-2x+8y-5 = 0

(b) 
x²-2x + 8y-5 = 0
(x-1)²-1 + 8y-5 = 0
(x-1)² + 8y-6 = 0
8y = -(x-1)² +6 
∴ y = -(¹/₈)(x-1)² + ¾ 

The locus of P is a parabola with :
vertex = (1,¾) , axis of symmetry : x=1
Further, it opens downwards.

設 P 點座標為 (x,y).

AB^2 = (x-2)^2+y^2
AC^2 = x^2+(y+2)^2
FP^2 = (y-3)^2

所以:
  [(x-2)^2+y^2] + [x^2+(y+2)^2] = 2(y-3)^2
化簡, 得
  (x-1)^2+8(y-3/4) = 0
也可寫成
  y = 3/4 - (x-1)^2/8

這是一個開口向下的拋物線, 
對稱軸 x = 1
頂點 (1,3/4)