✔ 最佳答案
C: x² + 2y² = 36
d(x² + 2y²)/dx = (d36/dx)
2x + 4yy' = 0
y' = -x/2y
設切點為 (x1, y1)。
在 (x1, y1) 上的切線斜率 = -x1/2y1
所求法線的斜率 = -1/(-x1/2y1) = 2y1/x1
(x1, y1) 在 C 上:
x1² + 2y1² = 36 ...... [1]
所求法線的斜率:
(y1 - 0)/(x1 - 1) = 2y1/x1
x1y1 = 2x1y1 - 2y1
x1y1 - 2y1 = 0
y1(x1 - 2) = 0
y1 = 0 或 x1 = 2
當 y1 = 0:
把 y1 = 0 代入 [1] 中:
x1² = 36
x1 = 6 或 x1 = -6
當 x1 = 2 :
把 x1 = 2 代入 [1] 中:
2² + 2y1² = 36
y1² = 16
y1 = 4 或 y1 = -4
所求的法線:
(y - 0)/(x - 6) = 2*0/6
y = 0
或:
(y - 0)/(x + 6) = 2*0/(-6)
y = 0
或:
(y - 4)/(x - 2) = 2*4/2
y - 4 = 4x - 8
4x - y - 4 = 0
或:
(y + 4)/(x - 2) = 2*(-4)/2
y + 4 = -4x + 8
4x + y - 4 = 0