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

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