导数及其应用

设切点为(x,y)
y = x² ...(#)
y' = 2x,这是曲线在切点处的斜率
切线斜率 = (x - 5)/(y - 3)
即2x = (x - 5)/(y - 3)
2x(y - 3) = x - 5
2xy - 6 = x - 5
2xy - x - 1 = 0,将(#)代入之
2x(x²) - x - 1 = 0
2x³ - x - 1 = 0
(x - 1)(2x² + 2x + 1) = 0
=> x = 1,2x² + 2x + 1 = 0无解
代x = 1到曲线方程得y = 1
所以直线方程为y = 2(1)(x - 1) + 1 即 y = 2x - 1
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y = x² + x - 2
y' = 2x + 1
y'(1) = 3
L₁为y = 3x - 3
L₂斜率为-1/3
设切点为(x,y)
符合y = x² + x - 2
y'(x) = 2x + 1 = -1/3
解得x = -2/3,所以y = -20/9
L₂为y = (-1/3)(x + 2/3) - 20/9 即 y = (-1/3)x - 22/9
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解L₁和L₂的交点：
{y = 3x - 3
{y = (-1/3)x - 22/9
得x = 1/6,y = -5/2
L₁与x轴交点为(1,0),L₂与x轴交点为(-22/3,0)
三个顶点为A：(1,0),B：(1/6,-5/2),C：(-22/3,0)
三角形的面积为：| 1 0 |
1/2 × | 1/6-5/2 |
| -22/3 0 |
| 1 0 |
= (1/2)| [(1)(-5/2) + (1/6)(0) + (-22/3)(0)] - [(0)(1/6) + (-5/2)(-22/3) + (0)(1)] |
= (1/2) × |-125/6|
= 125/12 = 10又5/12
A：(1,0),B：(1/6,-5/2),C：(-22/3,0)
如果不会上面那个方法,先求高：AB = √[(1 - 1/6)² + (0 + 5/2)²] = 5√10/6
再求底长：BC = √[(1/6 + 22/3)² + (-5/2 - 0)²] = 5√10/2
所以三角形面积= 1/2 * 底 * 高 = 1/2 * 5√10/2 * 5√10/6 = 125/12切线斜率 = (x - 5)/(y - 3)这个是不是应该改为（y-5）/（x-3）(y - 5)/(x - 3) = 2xy - 5 = 2x(x - 3)2x² - 6x + 5 = y2x² - 6x + 5 = x²x² - 6x + 5 = 0x = 5 or x = 1切点为(1，1) 或 (5，25)y = 2(1)(x - 1) + 1y = 2x - 1y = 2(5)(x - 5) + 25y = 10x - 25所以两条切线分别为y = 2x - 1 和 y = 10x - 25