过原点，倾斜角为θ的直线极坐标方程为什么加ρ∈R 过点（ρ1,θ1) 倾斜角为α的直线的极坐标方程如何推导?不...
\u8fc7\u6781\u70b9\uff0c \u503e\u659c\u89d2\u4e3a\u03b1 (1)\u03b8\uff1d\u03b1(\u03c1\u2208R) \u6216\u03b8\uff1d\u03b1\uff0b\u03c0\uff0c\u8fd9\u4e2a\u03b8\uff1d\u03b1\uff0b\u03c0\u8981\u600e\u4e48\u7406\u89e3\uff1f\uff081\uff09\u5c06\u66f2\u7ebf\u03c1 2 -6\u03c1cos\u03b8+5=0\u5316\u6210\u76f4\u89d2\u5750\u6807\u65b9\u7a0b\uff0c\u5f97\u5706C\uff1ax 2 +y 2 -6x+5=0 \u76f4\u7ebfl\u7684\u53c2\u6570\u65b9\u7a0b\u4e3a x=-1+tcos\u03b1 y=tsin\u03b1 \uff08t\u4e3a\u53c2\u6570\uff09\u5c06\u5176\u4ee3\u5165\u5706C\u65b9\u7a0b\uff0c\u5f97\uff08-1+tcos\u03b1\uff09 2 +\uff08tsin\u03b1\uff09 2 -6tsin\u03b1+5=0 \u6574\u7406\uff0c\u5f97t 2 -8tcos\u03b1+12=0 \u2235\u76f4\u7ebfl\u4e0e\u5706C\u6709\u516c\u5171\u70b9\uff0c \u2234\u25b3\u22650\uff0c\u537364cos 2 \u03b1-48\u22650\uff0c\u53ef\u5f97cos\u03b1\u2264- 3 2 \u6216cos\u03b1\u2265 3 2 \u2235\u03b1\u4e3a\u76f4\u7ebf\u7684\u503e\u659c\u89d2\uff0c\u5f97\u03b1\u2208[0\uff0c\u03c0\uff09 \u2234\u03b1\u7684\u53d6\u503c\u8303\u56f4\u4e3a[0\uff0c \u03c0 6 ]\u222a[ 5\u03c0 6 \uff0c\u03c0\uff09\uff082\uff09\u7531\u5706C\uff1ax 2 +y 2 -6x+5=0\u5316\u6210\u53c2\u6570\u65b9\u7a0b\uff0c\u5f97 x=3+2cos\u03b8 y=2sin\u03b8 \uff08\u03b8\u4e3a\u53c2\u6570\uff09 \u2235M\uff08x\uff0cy\uff09\u4e3a\u66f2\u7ebfC\u4e0a\u4efb\u610f\u4e00\u70b9\uff0c \u2234x+y=3+2cos\u03b8+2sin\u03b8=3+2 2 sin\uff08\u03b8+ \u03c0 4 \uff09 \u2235sin\uff08\u03b8+ \u03c0 4 \uff09\u2208[-1\uff0c1] \u22342 2 sin\uff08\u03b8+ \u03c0 4 \uff09\u2208[-2 2 \uff0c2 2 ]\uff0c\u53ef\u5f97x+y\u7684\u53d6\u503c\u8303\u56f4\u662f[3-2 2 \uff0c3+2 2 ]\uff0e
\u8fd9\u4e2a\u5e94\u8be5\u662f\u76f4\u63a5\u7684\u7ed3\u8bba\u3002
\u5de6\u8fb9\u03c1sin(\u03b1-\u03b8),\u662f\u6781\u70b9O\u5230\u76f4\u7ebfl\u7684\u8ddd\u79bb
\u53f3\u8fb9\u03c11*sin(\u03b1-\u03b81),\u4e5f\u662f\u6781\u70b9O\u5230\u76f4\u7ebfl\u7684\u8ddd\u79bb
\u6240\u4ee5\uff0c\u4e24\u8005\u76f8\u7b49
θ=α，ρ∈R。
若不附加说明ρ∈R，
则表示射线。
附
极坐标的点(ρ,θ)，
其中极径ρ≥0。
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