lulu810118002
幼苗
共回答了12个问题采纳率:83.3% 举报
设a^2=x1,a=y1,b^2=x2,b=y2
则a^2sinθ+acosθ-(π/4)=0,b^2sinθ+bcosθ-(π/4)=0可得x1sinθ+y1cosθ-(π/4)=0,x2sinθ+y2cosθ-(π/4)=0
点A(a^2,a),B(b^2,b)等价于(x1,y1),(x2,y2)
则点(x1,y1),(x2,y2)都在直线xsinθ+ycosθ-(π/4)=0上
即点A,B在直线xsinθ+ycosθ-(π/4)=0上
因为点(0,0)到直线xsinθ+ycosθ-(π/4)=0的距离是(π/4)/√(sin^2θ+cos^2θ)=(π/4)
1年前
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