求过曲线y=-x^2+1上一点,使得过该点的切线与这条曲线及x轴y轴再第一象限所围成面积最小

Question

求过曲线y=-x^2+1上一点,使得过该点的切线与这条曲线及x轴y轴再第一象限所围成面积最小

Answer 1

y = -x² + 1 = 0,x = 1 (舍去x = -1 < 0)
曲线与x轴交于A(1,0)
设切点P(p,1 - p²),0 < p < 1
y = 1 - x²,y' = -2x
过点P的切线:y - (1 - p²) = -2p(x - p)
y = -2px + 1 + p²
取x = 0,y = 1 + p²
取y = 0,x = (1 + p²)/(2p)
切线与x轴,y轴交于B((1 + p²)/(2p),0),C(0,1 + p²)
三角形OBC面积S1 = (1/2)OB*OC = (1 + p²)²/(4p)
曲线与x轴,y轴于第一象限所围成面积S2 = ∫₀¹(-x² + 1)dx = (-x³/3 + x)|₀¹
= 1 - 1/3 = 2/3
切线与这条曲线及x轴y轴于第一象限所围成面积S = S1 - S2
= (1 + p²)²/(4p) - 2/3
求导S' = (1/4)[2(1 + p²)*2p*p - (1 + p²)²]/p²
= (1 + p²)[4p² - (1 + p²)]/(4p²) = 0
3p² = 1
p = 1/√3 (舍去p = -1/√3 < 0)
P(1/√3,2/3)