已知两个不同的点A(-cosα,3sin^2α)B(根号3sinα,cos^2α),α∈[π/2,π],求直线AB的斜率和倾斜角的范围

Answer 1

直线AB的斜率：k=(cos²α-3sin²α)/(√3sinα+cosα)
=(cosα-√3sinα)(cosα+√3sinα)/(cosα+√3sinα)
=2(1/2cosα-√3/2sinα)
=2sin(π/6-α)
=-2sin(α-π/6)
-2√3≤k≤-2√3/3
-2√3≤tanx≤-2√3/3
arctan(-2√3)≤x≤arctan(-2√3/3)