A.Maths - 軌跡問題

Wai Chi L

2008-05-29 3:44 am

1. 已知點P在圎C: x² + y² + 4x - 6y + 9 = 0 上移動 , A(1, 2)為一固定點。若AP的中點為M , 試求點M的軌跡方程。

2. 已知兩點 A(-2, 1)和 B(2, 0)。一動點P移動時 , PA的斜率恒等於PB的斜率的2倍。試求點P的軌跡方程。

更新1:

Answer : 1. 2x² + 2y² + 2x - 10y + 11 = 0 2. xy + x + 6y - 2 = 0

回答 (1)

用戶10012

2008-05-29 4:13 am

✔ 最佳答案

Q.1 Let M be (h,k) and P be (x,y). Then h=(x+1)/2 and k=(y+2)/2.
That is x=2h-1 and y=2k-2.
Since P is on the circle, therefore, (x,y) satisfies the circle equation, that is
(2h-1)^2 + (2k-2)^2 + 4(2h-1) -6(2k-2) + 9 = 0
4h^2 -4h + 1 + 4k^2 -8k + 4 + 8h - 4 -6k + 12 +9 = 0
4h^2 + 4k^2 + 4h -14k + 26 = 0
2h^2 + 2k^2 + 2h -7k +13 = 0
or 2x^2 + 2y^2 +2x -7y +13 = 0 is the locus of M.
Q.2 Slope of PA = (y-1)/(x +2), slope of PB= y/(x-2)
Therefore, (y-1)/(x+2)= 2y/(x-2)
(y-1)(x-2) = 2y(x+2)
xy -x -2y +3 = 2xy +4y
xy + x +6y -3 = 0 is the locus of P.

2008-05-28 20:24:04 補充：
Correction: Should be 4h^2 -4h + 1 +4k^2 -8k +4 +8h - 4 -12k +12 + 9 = 0.
Answer is therefore 2h^2 +2k^2 +2h -10k +11 = 0.
Correction: Should be xy - x -2y +2 = 2xy +4y, therefore, xy +x +6y -2=0

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