求在點(1,1)正切於X^2+XY+Y^2=3的直線方程式
2014-05-27 5:06 am
An equation of the tangent line to the curve X^2+XY+Y^2=3 at the point (1,1) is ? .
回答 (6)
2014-05-27 6:10 am
✔ 最佳答案
x² + xy + y² = 3(d/dx) (x² + xy + y²) = (d/dx) 3
2x + xy' + y + 2yy' = 0
xy' + 2yy' = -2x - y
(x + 2y)y' = -(2x + y)
y' = -(2x + y)/(x + 2y)
Slope of the tangent at (1, 1)
= y' at (1, 1)
= -(2 + 1)/(1 + 2)
= -1
Equation of the tangent :
(y - 1) = (-1)(x - 1)
y - 1 = -x + 1
x + y - 2 = 0
參考: 胡雪
2014-06-21 10:32 pm
2014-06-20 9:45 pm
2014-06-11 12:47 am
2014-06-03 2:58 pm
2014-05-30 4:15 pm
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原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20140526000010KK06981