点P的极坐标为P(ρ1,θ1),直线L过点P,且与极轴夹角为α,求直线的极坐标方程,有解析我看不懂,解析:如图,由正弦定理|OM|/sin∠OPM=|OP|/sin∠OMP ,即ρ/sin[π-(α-θ1)]=ρ1/sin(α-θ)即ρsin(α-θ)=ρ1sin(α-θ1),显

点P的极坐标为P(ρ1,θ1),直线L过点P,且与极轴夹角为α,求直线的极坐标方程,有解析我看不懂,解析:如图,由正弦定理|OM|/sin∠OPM=|OP|/sin∠OMP ,即ρ/sin[π-(α-θ1)]=ρ1/sin(α-θ)即ρsin(α-θ)=ρ1sin(α-θ1),显
点P的极坐标为P(ρ1,θ1),直线L过点P,且与极轴夹角为α,求直线的极坐标方程,有解析我看不懂,
解析:如图,由正弦定理|OM|/sin∠OPM=|OP|/sin∠OMP ,即ρ/sin[π-(α-θ1)]=ρ1/sin(α-θ)
即ρsin(α-θ)=ρ1sin(α-θ1),显然,点P的坐标(ρ1,θ1)是方程
{x=ρcosθ
{y=ρsinθ 的解,所以方程ρsin(α-θ)=ρ1sin(α-θ1) 是直线L的极坐标方程.【为什么说点P的坐标(ρ1,θ1)是方程的解?还有为什么因为点P是方程x、y的解就说明ρsin(α-θ)=ρ1sin(α-θ1)是直线L的极坐标方程呢?】
图

点P的极坐标为P(ρ1,θ1),直线L过点P,且与极轴夹角为α,求直线的极坐标方程,有解析我看不懂,解析:如图,由正弦定理|OM|/sin∠OPM=|OP|/sin∠OMP ,即ρ/sin[π-(α-θ1)]=ρ1/sin(α-θ)即ρsin(α-θ)=ρ1sin(α-θ1),显
点P是方程x、y的解就说明ρsin(α-θ)=ρ1sin(α-θ1)是直线L
∵P是不定点(也就是任意P点都一定在L上)
∴不定点所在的极坐标方程就是直线的极坐标方程

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