A.Maths - 軌跡問題

Wai Chi L

2008-05-28 7:49 am

2. 一動點P移動時,其與點(2, -2) 的距離恒等於其與直線 x- y = 0 的距離。求點P的軌跡方程。

３. 已知A(-2, 3) 為一固定點及P為一動點。若AP的中點恒位於直線
L: 3x - 4y + 6 = 0 上　,　求點P的軌跡方程。

回答 (1)

用戶10012

2008-05-28 4:03 pm

✔ 最佳答案

Q.2
Let P be (x,y). Distance from P to point (2,-2) = sqrt[(x-2)^2 + (y+2)^2].
Distance from P to line x-y=0 is (x-y)/sqrt2.
Therefore, sqrt[(x-2)^2 + (y+2)^2] = (x-y)/sqrt2
(x-2)^2 + (y+2)^2 = (x-y)^2/2
x^2 -4x + 4 + y^2 +4y + 4 = 1/4(x^2+-y^2 -2xy)
3x^2 + 3y^2 + 2xy -16x +16y +32 = 0 is the locus of P.
Q.3
Mid-point of AP is: [(x-2)/2, (y+3)/2]
Since this point is always on line L. That is
3(x-2)/2 -4(y+3)/2 + = 0
3(x -2) -4(y+3) +2 = 0
3x -6 -4y -12 +2 = 0
3x -4y -16 = 0 is the locus of P.

2008-05-28 08:09:10 補充：
Correction: 3(x-2)/2 -4(y+3)/2 +6=0, that is 3x -4y -6 = 0.

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