A.Maths - 軌跡及參數方程

Wai Chi L

2008-05-23 5:35 am

1. 一動點至點 A(2, 3)的距離等於至點 B(3, 1) 的距離的2倍 , 求這點的軌跡方程。

2. 點A和B的坐標分別為(0, 2)和(9, 8)。一動點P移動時 , ∠APB恒為直角。求點P的軌跡方程。

回答 (1)

用戶10012

2008-05-23 5:49 am

✔ 最佳答案

1. Let the moving point be (x,y).
Distance between this point and A = sqrt[(x-2)^2 + (y-3)^2]
Distance between this point and B =sqrt[(x-3)^2 + (y-1)^2]
Therefore, (x-2)^2 + (y-3)^2 = 4[(x-3)^2 + (y-1)^2]
x^2 -4x + 4 + y^2 -6y + 9 = 4x^2 -24x + 36 + 4y^2 -8y + 4
Therefore, locus is : 3x^2 + 3y^2 -20x -2y + 27 = 0.
2. Let (x,y) be the moving point, P.
Slope of line AP = y-2/x
Slope of line BP = (y-8)/(x-9)
Since angle APB is a right angle, therefore slope of AP x slope of BP = -1.
That is (y-2)(y-8)/x(x-9) = -1
y^2 -10y + 16 = -x^2 + 9x
Therefore locus is: x^2 + y^2 -9x -10y +16 = 0.

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