直线x+2y-4=0绕x轴旋转的圆锥面面方程怎么求? 急！若将直线x+2y-4=0绕着它与x轴的交点按逆时针方向旋...
\u76f4\u7ebfx+2y-4=0\u7ed5x\u8f74\u65cb\u8f6c\u7684\u5706\u9525\u9762\u9762\u65b9\u7a0b\u600e\u4e48\u6c42?\u8bbe\u65cb\u8f6c\u9762\u4e0a\u4efb\u610f\u4e00\u70b9\u4e3ap\uff08x,y,z)\uff0c\u5b83\u662f\u7531\u76f4\u7ebf\u4e0a\u7684\u70b9p0(2y,y,1/2(y+1))\u65cb\u8f6c\u8fc7\u6765\u7684\u3002p\u5230y\u8f74\u7684\u8ddd\u79bb\uff0c\u5e94\u4e0ep0\u5230y\u8f74\u7684\u8ddd\u79bb\u76f8\u7b49\u3002\u5373x^2+z^2=(2y)^2+[1/2(y+1)]^2,\u65cb\u8f6c\u9762\u7684\u65b9\u7a0b\u4e3a\uff1a
4x²-17y²+4z²-2y=1
\u6253\u5b57\u5199\u8fc7\u7a0b\u592a\u9ebb\u70e6 \u76f4\u63a5\u5199\u7684\u8349\u7a3f
\u8fc7\u7a0b\u89c1\u56fe\u7247
\u6700\u7ec8\u7b54\u6848\u4e3ax-3y-4=0
则满足x=x₁，(y,z)在以|y₁|为半径的圆上，故y²+z²=y₁²
又(x₁,y₁)满足x₁+2y₁-4=0，y₁=(4-x₁)/2=(4-x)/2
代入得：y²+z²=y₁²=[(4-x)/2]²
化简得圆锥面方程为：(x-4)²=4(y²+z²)
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