Question

在极坐标系中,直线psin(θ+π/4)=2被圆p=4截得的弦长为?

Answer 1

PSIN(θ+π/4)=P(SINθ*COSπ/4+COSθ*SINπ/4)=√2/2*PSINθ+√2/2*PCOSθ

=√2/2(X+Y)=2

=> X+Y=2√2 =>X+Y-2√2=0

P=4 => P²=16 => X²+Y²=16 圆心(0,0),半径R=4.

圆心到直线距离:D=|-2√2|/√(1+1)=2

设弦长为L,根据垂径定理:(L/2)²=R²-D²=4²-2²=16-4=12

=>L=4√3

因此,答案为4√3