求在點(1,1)正切於X^2+XY+Y^2=3的直線方程式

👨🏻‍🏫 學術台數學

[匿名]

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)

胡雪8°

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

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[匿名]

2014-06-21 10:32 pm

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[匿名]

2014-06-20 9:45 pm

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[匿名]

2014-06-11 12:47 am