设AB=m,AB与x轴夹角=α
B(k,1/k)
A(k+mcosα,1/k+msinα)
C(k+msinα, 1/k-mcosα)
求m的最小值
解:
(k+mcosα)(1/k+msinα)=1
(k+msinα)(1/k-mcosα)=1
kmsinα+mcosα/k+m²sinαcosα=0
-kmcosα+msinα/k-m²sinαcosα=0
-k²msinα-mcosα=-k²mcosα+msinα
(sinα+cosα)/(cosα-sinα)=k²
(sinα/k-kcosα)/(sinαcosα)=m
m²=(sin²α/k²+k²cos²α-2sinαcosα)/sin²αcos²α
=[sin²α(cosα-sinα)/(sinα+cosα)+cos²α(sinα+cosα)/(cosα-sinα)-2sinαcosα]/sin²αcos²α
=[sin²α(cosα-sinα)²+cos²α(sinα+cosα)²-2sinαcosα(cos²α-sin²α)]/(cos²α-sin²α)*sin²αcos²α
=[sin²α(1-sin2α))+cos²α(1+sin2α)-sin2αcos2α]/cos2α*(1/4)sin2α
=4/cos2αsin2α
=8/sin4α
α=π/8的时候,sin4α取到最大值1,8/sin4α取到最小值8,S△ABC=m²/2=4