已知定点A(4,0)和圆M:x^2+y^2=9/4,设B是圆M上的动点,点P满足AP向量=2PB向量,(1)求点P的轨迹方程.(3)将(1)所得的点P按向量a=(2/3,3)平移得轨迹C,从轨迹C外一点R(x0,y0)向轨迹C作切线RT,T是切点,且R
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![已知定点A(4,0)和圆M:x^2+y^2=9/4,设B是圆M上的动点,点P满足AP向量=2PB向量,(1)求点P的轨迹方程.(3)将(1)所得的点P按向量a=(2/3,3)平移得轨迹C,从轨迹C外一点R(x0,y0)向轨迹C作切线RT,T是切点,且R](/uploads/image/z/8554087-55-7.jpg?t=%E5%B7%B2%E7%9F%A5%E5%AE%9A%E7%82%B9A%284%2C0%29%E5%92%8C%E5%9C%86M%3Ax%5E2%2By%5E2%3D9%2F4%2C%E8%AE%BEB%E6%98%AF%E5%9C%86M%E4%B8%8A%E7%9A%84%E5%8A%A8%E7%82%B9%2C%E7%82%B9P%E6%BB%A1%E8%B6%B3AP%E5%90%91%E9%87%8F%3D2PB%E5%90%91%E9%87%8F%2C%EF%BC%881%EF%BC%89%E6%B1%82%E7%82%B9P%E7%9A%84%E8%BD%A8%E8%BF%B9%E6%96%B9%E7%A8%8B.%EF%BC%883%EF%BC%89%E5%B0%86%EF%BC%881%EF%BC%89%E6%89%80%E5%BE%97%E7%9A%84%E7%82%B9P%E6%8C%89%E5%90%91%E9%87%8Fa%3D%282%2F3%2C3%29%E5%B9%B3%E7%A7%BB%E5%BE%97%E8%BD%A8%E8%BF%B9C%2C%E4%BB%8E%E8%BD%A8%E8%BF%B9C%E5%A4%96%E4%B8%80%E7%82%B9R%28x0%2Cy0%29%E5%90%91%E8%BD%A8%E8%BF%B9C%E4%BD%9C%E5%88%87%E7%BA%BFRT%2CT%E6%98%AF%E5%88%87%E7%82%B9%2C%E4%B8%94R)
已知定点A(4,0)和圆M:x^2+y^2=9/4,设B是圆M上的动点,点P满足AP向量=2PB向量,(1)求点P的轨迹方程.(3)将(1)所得的点P按向量a=(2/3,3)平移得轨迹C,从轨迹C外一点R(x0,y0)向轨迹C作切线RT,T是切点,且R
已知定点A(4,0)和圆M:x^2+y^2=9/4,设B是圆M上的动点,点P满足AP向量=2PB向量,
(1)求点P的轨迹方程.(3)将(1)所得的点P按向量a=(2/3,3)平移得轨迹C,从轨迹C外一点R(x0,y0)向轨迹C作切线RT,T是切点,且RT=RO,O为原点,求RT的最小值.
只要第三问,
已知定点A(4,0)和圆M:x^2+y^2=9/4,设B是圆M上的动点,点P满足AP向量=2PB向量,(1)求点P的轨迹方程.(3)将(1)所得的点P按向量a=(2/3,3)平移得轨迹C,从轨迹C外一点R(x0,y0)向轨迹C作切线RT,T是切点,且R
只要第三问,没第一问怎么做第三问呢?
设P点坐标(x,y),B点坐标(3cosa/2,3sina/2),则:AP=(x,y)-(4,0)=(x-4,y)
PB=(3cosa/2,3sina/2)-(x,y)=(3cosa/2-x,3sina/2-y),而:AP=2PB,故:
(x-4,y)=2(3cosa/2-x,3sina/2-y),即:x-4=3cosa-2x,y=3cosa-2y
即:3x-4=3cosa,3y=3sina,故:(3x-4)^2+9y^2=9,即:(x-4/3)^2+y^2=1
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按照向量a=(2/3,3)平移后,轨迹C还是一个圆,圆心Q:(4/3+2/3,3)=(2,3),半径:1
故C:(x-2)^2+(y-3)^2=1,|RT|^2=|RQ|^2-|QT|^2=(x0-2)^2+(y0-3)^2-1,而:
|RO|^2=x0^2+y0^2,故:(x0-2)^2+(y0-3)^2-1=x0^2+y0^2,即:2x0+3y0-6=0
当圆C的直径与直线2x0+3y0-6=0垂直时,|RT|=|RO|最小,此时圆C直径所在直线方程:
y-3=3(x-2)/2,即:y=3x/2,与直线2x+3y-6=0联立可得交点坐标为(12/13,18/13)
故RT的最小值为:sqrt((12/13)^2+(18/13)^2)=6/sqrt(13)=6sqrt(13)/13