急 ! F5 數 Locus 9q4

11.
(a)
Locus of P are 2 straight lines parallel to x = 3, with distance 1 from x = 3.
(b)
x = 3
i.e. x - 3 = 0
Therefore, the locus of P(x, y):
|x - 3|/(1^2)^(1/2) = 1
x - 3 = +/- 1
i.e.
x - 3 = -1 or x - 3 = 1
x = 2 or x = 4

14.
Locus of P is a horizontal parabola that
passes through the vertex which is the mid-point of DE,
and ends at B and H, with the axis of symmetry DF.

題目說 P與E和線AG的距離相等。
不妨設 E = (0, 0)
則A = (-3, 3)
G = (-3, -3)
AG 的方程: x = -3

而P和E的距離 = P 和 線AG的距離:
[(x - 0)^2 + (y - 0)^2]^(1/2) = |x + 3|/[(1)^(2)]^(1/2)
i.e.
x^2 + y^2 = (x + 3)^2
x^2 + y^2 = x^2 + 6x + 9
x = (1/6)y^2 - 3/2, 明顯地是一個橫向的拋物線
拋物線的對稱軸: y = 0, 也就是線DF。

因為P只在正方形ACIG內移動,
所以
-3 ≤ y ≤ 3 和 -3 ≤ x ≤ 3
因為x = (1/6)y^2 - 3/2
-> y^2 = 6x + 9
所以
-3 ≤ (6x + 9)^(1/2) ≤ 3
0 ≤ 6x + 9 ≤ 9
-9 ≤ 6x ≤ 0
-3/2 ≤ x ≤ 0
即, -3 ≤ y ≤ 3 和 -3/2 ≤ x ≤ 0

當 y = -3, x = 3/2 - 3/2 = 0
當 y = 3, x = 3/2 - 3/2 = 0
當 x = -3/2, y = 0
也就是 P的路徑 穿過 (0, 3) ,(0, -3) 和 頂點(-3/2, 0)
即 P的路徑 穿過點B、點H 和 DE的中點。
而因為 -3 ≤ y ≤ 3, P的路徑兩端以點B和點H為終點。

(簡單來說,
因為AB = BE, GH = EH
所以P的路徑一定經過點B和點H。
設DE的中點為M,
則DM = ME(當中,DM是M與線AG的距離)
所以 P的路徑經過DE的中點。)

其實這題…還真不知道該怎麼回答才是最好的= =..
對了, 圖在這裡:

圖片參考：http://imgcld.yimg.com/8/n/HA05426997/o/20121214132220.jpg