附加數學問題(10點)

設 P 的坐標為 (x, y)。

PM
= |2x - y|/√(22 + 12)
= |2x - y|/√5

PN
= |y|

PM = PN
|2x - y|/√5 = |b|
2x - y = y√5 或 2x - y = -y√5

P 點的軌跡是:
2x - (1 + √5)y = 0 或 2x - (1 - √5)y = 0
=

001 看漏了題目：
PM = |2x - y|/√5
PN = |y|
PM = (√5)PN
|2x - y|/√5 = (√5)|y|
2x - y = 5y 或 2x - y = -5y
2x - 6y = 0 或 2x + 4y = 0
P 點的軌跡是: x - 3y = 0 或 x + 2y = 0

2009-05-01 23:42:19 補充：
002 和 003 用開方法，但沒有計算到答案。
PM^2 = (√5PN)^2
[(2x - y)/(√5)]^2 = [(√5)y]^2
(4x^2 - 4xy + y^2)/5 = 5y^2
4x^2 - 4xy + y^2 = 25y^2
4x^2 - 4xy - 24y^2 = 0
x^2 - xy - 6y^2 = 0
(x - 3y)(x + 2y) = 0
P 點的軌跡是: x - 3y = 0 或 x + 2y = 0

Let P be (x,y)

PM=絕對值-2x+y / square root 5

PN=絕對值y

Because PM=√5PN(我用:代表絕對值)

So :-2x+y /sqrt 5:=sqrt 5 :y:

square both side

(-2x+y)^2=25y^2

4x^2-4xy+y^2-25y^2=0

So the locus of P is 4x^2-24y^2-4xy=0(or x^2-6y^2-xy=0)

http://img216.imageshack.us/my.php?image=locus.jpg

Let P be ( x,y).
Distance from P to line y - 2x = 0 is
( y - 2x)/sqrt(1 + 2^2) = (y -2x)/sqrt 5 = PM.
Distance from P to x - axis = y.
since PM^2 =5PN^2,
that is (y-2x)^2/5 = y^2
y^2 + 4x^2 - 4xy = 5y^2
4x^2 - 4y^2 - 4xy = 0
x^2 - y^2 - xy = 0 is the locus of P.