曲线函数=x∧2 -ax+a,曲线外一点A(1,0)引曲线的两条切线,它们的倾斜角互补,a的取值

庄财主2022-10-04 11:39:541条回答

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albert784290 共回答了22个问题 | 采纳率95.5%
y' = 2x - a
设切点P(p,p² - ap + a)
切线斜率k = 2p - a
斜率方程:y - (p² - ap + a) = (2p -a)(x - p)
过点A:0 - (p² - ap + a) = (2p -a)(1 - p)
p² - 2p = 0
p = 0,p = 2
两条切线斜率:k1 = -a,k2 = 4 - a
倾斜角互补,k1 = -k2
-a = a - 4
a = 2
1年前

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