CE A.Maths (locus)

Two points A(-3,2) and B(5,2) are given. If P is a movable point such that the perimeter of triangleABP is 18, find the equation of locus of P.
Let the point P(x,y)
AB=8
PA=√[(x+3)^2+(y-2)^2]
PB=√[(x-5)^2+(y-2)^2]
AB+PA+PB=18
8+√[(x+3)^2+(y-2)^2]+√[(x-5)^2+(y-2)^2]=18
√[(x+3)^2+(y-2)^2]+√[(x-5)^2+(y-2)^2]=10
√[(x+3)^2+(y-2)^2]=10-√[(x-5)^2+(y-2)^2]
(x+3)^2+(y-2)^2=100-20√[(x-5)^2+(y-2)^2]+(x-5)^2+(y-2)^2
16x-16=100-20√[(x-5)^2+(y-2)^2]
20√[(x-5)^2+(y-2)^2]=116-16x
5√[(x-5)^2+(y-2)^2]=29-4x
25[(x-5)^2+(y-2)^2]=841-232x+16x^2
25x^2+25y^2-250x-100y+725=841-232x+16x^2
9x^2+25y^2-18x-100y-116=0
9(x-1)^2+25(y-2)^2=225
(x-1)^2/25+(y-2)^2/9=1


2007-11-04 13:11:43 補充：
上面個位計錯數

Let P be (x, y)
Length of AB = 5-(-3) =8
Hence perimeter of triangle ABP is
8+ sqrt[(x+3)^2 + (y-2)^2] + sqrt[(x-5)^2 + (y-2)^2] = 18
sqrt[(x-5)^2 + (y-2)^2] = 10 - sqrt[(x+3)^2 + (y-2)^2]
(x-5)^2 + (y-2)^2 = 100 + (x+3)^2 + (y-2)^2 -10sqrt[(x+3)^2 + (y-2)^2]
-16x-84 = -10sqrt[(x+3)^2 + (y-2)^2]
256x^2 +2688x + 7056 = 100(x^2 + 6x + 9 +y^2 - 4y + 4)
39x^2 + 522x +1439-25y^2 + 100y = 0
39x^2 - 25y^2 + 522x + 100y + 1439 =0