1 MATH MC

Let P(x,y) be any point lies on the required locus.

PA=√[(x-0)^2+(y-0)^2]=√[x^2+y^2]
PB=√[(x-0)^2+(y-2)^2]=√[x^2+(y-2)^2]

PA+PB=4
√[x^2+y^2] + √[x^2+(y-2)^2] = 4
√[x^2+y^2] = 4 - √[x^2+(y-2)^2]

Take square both side,
x^2+y^2 = 16 - 8√[x^2+(y-2)^2] + [x^2+(y-2)^2]

Simplify it and we have
2√[x^2+(y-2)^2] = 5 - y

Taking square again, we have
4[x^2+(y-2)^2] = (5-y)^2

Simplify it and finally we get
4x^2 + 3y^2 - 6y - 9 = 0

PS. Do your question have some problem?
I have checked my work many times,
but I still cannot find any mistake.



2008-12-12 15:57:53 補充：
Maybe the answer is E...=.="

Let point P(x,y)

THen

PA+PB=4

√(x^2+y^2)+√[x^2+(y-2)^2)]=4

(x^2+y^2)=[4-√[x^2+(y-2)^2)]]^2

x^2+y^2=16-8√[x^2+(y-2)^2)]+x^2+(y-2)^2

y=5-2√[x^2+(y-2)^2)]

2√[x^2+(y-2)^2)]=5-y

4(x^2+y^2-4y+4)=25-10y+y^2

4x^2+3y^2-6y-9=0

Let the point P be (x, y).
AP^2 = (x - 0)^2 + (y -0)^2 = x^2 + y^2.
PB^2 = ( x- 0)^2 + (y -2)^2 = x^2 + (y -2)^2 .
Since PA + PB = 4
PA = 4 - PB
PA^2 = 16 + PB^2 - 8PB
x^2 + y^2 = 16 + x^2 + y^2 + 4 - 4y - 8PB
8PB = 20 - 4y
2PB = 5 - y
4PB^2 = (5 - y)^2 = 25 + y^2 - 10y
4(x^2 + y^2 + 4 - 4y) = 25 + y^2 - 10y
4x^2 + 4y^2 + 16 - 16y - 25 - y^2 + 10y = 0
4x^2 + 3y^2 - 6y - 9 = 0

2008-12-12 15:39:02 補充：
Either you or me made a mistake, anyway, this is the method to find the locus of P.