- 过点M(
- 过点M(-2,0)作直线L交双曲线x^2-y^2=1于A,B两点,已知向量OP=向量OA+向量OB,求点P的轨迹方程
- 设直线L的方程为 y = k(x+2)
与双曲线方程 x² - y² = 1 联立,消去 y ,并整理得
(1-k²)x² - 4k²x - (4k²+1) = 0
设 A(x1, y1) , B(x2, y2),
则 x1 + x2 = 4k²/(1-k²)
y1 + y2 = k(x1 + 2) + k(x2 + 2) = k(x1 + x2 + 4) = 4k/(1-k²)
设 P(x, y),则因为 OP = OA + OB
所以 x = x1 + x2 = 4k²/(1-k²) ............(1)
y = y1 + y2 = 4k/(1-k²) ............(2)
(1) / (2) , 得 k = x/y
代入 (2) , 得 y = 4(x/y)/[1-(x/y)²]
整理,得 (x+2)² - y² = 4