圆锥曲线提问已知动点P到点A(-2,0)与点B(2,0)的斜率之积为-1/4,点P的轨迹为曲线C1.求曲线C的方程2.若点Q为曲线C上的一点,直线AQ,BQ与直线x=4分别交于M,N两点,直线BM与椭圆的交点为D,求证A,D,N三点

Question

圆锥曲线提问
已知动点P到点A(-2,0)与点B(2,0)的斜率之积为-1/4,点P的轨迹为曲线C 1.求曲线C的方程 2.若点Q为曲线C上的一点,直线AQ,BQ与直线x=4分别交于M,N两点,直线BM与椭圆的交点为D,求证A,D,N三点共线.

Answer 1

1.
P(x, y)
PA的斜率a = (y - 0)/(x + 2) = y/(x + 2)
PB的斜率b = (y - 0)/(x- 2) = y/(x - 2)
ab = y²/(x² - 4) = -1/4
x²/4 + y² = 1
2.
Q(u, v), u²/4 + v² = 1, 4v² = 4 - u² (i)
AQ的方程: (y - 0)/(v - 0) = (x + 2)/(u + 2), 取x = 4, y = 6v/(u + 2), M(4, 6v/(u + 2))
BQ的方程: (y - 0)/(v - 0) = (x - 2)/(u - 2), 取x = 4, y = 4v/(u - 2), N(4, 2v/(u - 2))
BM的方程:
(y - 0)/[6v/(u + 2) - 0] = (x - 2)/(4 - 2)
x = 2 + (u + 2)y/(3v)
代入x²/4 + y² = 1, 得y = -12v(u + 2)/[(u + 2)² + 36v²] (舍去y = 0, 此为点B)
x = 2 + (u + 2)y/(3v) = 2 - 4(u + 2)²/[(u + 2)² + 36v²]
D(2 - 4(u + 2)²/[(u + 2)² + 36v²], -12v(u + 2)/[(u + 2)² + 36v²])
AN的方程:
(y - 0)/[2v/(u - 2) - 0] = (x + 2)/(4 + 2)
3(u - 2)y = v(x + 2)
代入D的坐标:
-36v(u² - 4)/[(u + 2)² + 36v²] = v{4 - 4(u + 2)²/[(u + 2)² + 36v²]}
整理得: -9(u² - 4)/[(u + 2)² + 36v²] = 1 - (u + 2)²/[(u + 2)² + 36v²]
(u + 2)²/[(u + 2)² + 36v²] - 9(u² - 4)/[(u + 2)² + 36v²] = 1
(u + 2)² - 9(u² - 4) = (u + 2)² + 36v²
由(i), 右边 = (u + 2)² + 9(4 - u²) = (u + 2)² - 9(u² - 4)
两边相等,即A, D, N共线