作业帮点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
来源：学生作业帮助网 编辑：作业帮 时间：2024/11/15 12:07:24
![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](/uploads/image/z/7154421-69-1.jpg?t=%E7%82%B9P%E7%9A%84%E6%9E%81%E5%9D%90%E6%A0%87%E4%B8%BAP%28%CF%811%2C%CE%B81%29%2C%E7%9B%B4%E7%BA%BFL%E8%BF%87%E7%82%B9P%2C%E4%B8%94%E4%B8%8E%E6%9E%81%E8%BD%B4%E5%A4%B9%E8%A7%92%E4%B8%BA%CE%B1%2C%E6%B1%82%E7%9B%B4%E7%BA%BF%E7%9A%84%E6%9E%81%E5%9D%90%E6%A0%87%E6%96%B9%E7%A8%8B%2C%E6%9C%89%E8%A7%A3%E6%9E%90%E6%88%91%E7%9C%8B%E4%B8%8D%E6%87%82%2C%E8%A7%A3%E6%9E%90%3A%E5%A6%82%E5%9B%BE%2C%E7%94%B1%E6%AD%A3%E5%BC%A6%E5%AE%9A%E7%90%86%7COM%7C%2Fsin%E2%88%A0OPM%3D%7COP%7C%2Fsin%E2%88%A0OMP+%2C%E5%8D%B3%CF%81%2Fsin%5B%CF%80-%28%CE%B1-%CE%B81%29%5D%3D%CF%811%2Fsin%28%CE%B1-%CE%B8%29%E5%8D%B3%CF%81sin%28%CE%B1-%CE%B8%29%3D%CF%811sin%28%CE%B1-%CE%B81%29%2C%E6%98%BE)
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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