已知平面上一定点C(2,0)和直线l:x=8,P为该平面上一动点,作PQ⊥l,垂足为Q,且(PC向量+1/2PQ向量)•(PC向量-1/2PQ向量)=0.(1)求动点P的轨迹方程.(2)若EF为圆N:x^2+(y-1)^2=1的任一条直线,求PE向
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![已知平面上一定点C(2,0)和直线l:x=8,P为该平面上一动点,作PQ⊥l,垂足为Q,且(PC向量+1/2PQ向量)•(PC向量-1/2PQ向量)=0.(1)求动点P的轨迹方程.(2)若EF为圆N:x^2+(y-1)^2=1的任一条直线,求PE向](/uploads/image/z/11492018-26-8.jpg?t=%E5%B7%B2%E7%9F%A5%E5%B9%B3%E9%9D%A2%E4%B8%8A%E4%B8%80%E5%AE%9A%E7%82%B9C%282%2C0%29%E5%92%8C%E7%9B%B4%E7%BA%BFl%3Ax%3D8%2CP%E4%B8%BA%E8%AF%A5%E5%B9%B3%E9%9D%A2%E4%B8%8A%E4%B8%80%E5%8A%A8%E7%82%B9%2C%E4%BD%9CPQ%E2%8A%A5l%2C%E5%9E%82%E8%B6%B3%E4%B8%BAQ%2C%E4%B8%94%28PC%E5%90%91%E9%87%8F%2B1%2F2PQ%E5%90%91%E9%87%8F%EF%BC%89%26%238226%3B%EF%BC%88PC%E5%90%91%E9%87%8F-1%2F2PQ%E5%90%91%E9%87%8F%EF%BC%89%3D0.%EF%BC%881%EF%BC%89%E6%B1%82%E5%8A%A8%E7%82%B9P%E7%9A%84%E8%BD%A8%E8%BF%B9%E6%96%B9%E7%A8%8B.%EF%BC%882%EF%BC%89%E8%8B%A5EF%E4%B8%BA%E5%9C%86N%EF%BC%9Ax%5E2%2B%28y-1%29%5E2%3D1%E7%9A%84%E4%BB%BB%E4%B8%80%E6%9D%A1%E7%9B%B4%E7%BA%BF%2C%E6%B1%82PE%E5%90%91)
已知平面上一定点C(2,0)和直线l:x=8,P为该平面上一动点,作PQ⊥l,垂足为Q,且(PC向量+1/2PQ向量)•(PC向量-1/2PQ向量)=0.(1)求动点P的轨迹方程.(2)若EF为圆N:x^2+(y-1)^2=1的任一条直线,求PE向
已知平面上一定点C(2,0)和直线l:x=8,P为该平面上一动点,作PQ⊥l,垂足为Q,且(PC向量+1/2PQ向量)•(PC向量-1/2PQ向量)=0.
(1)求动点P的轨迹方程.
(2)若EF为圆N:x^2+(y-1)^2=1的任一条直线,求PE向量•PF向量的最值.
已知平面上一定点C(2,0)和直线l:x=8,P为该平面上一动点,作PQ⊥l,垂足为Q,且(PC向量+1/2PQ向量)•(PC向量-1/2PQ向量)=0.(1)求动点P的轨迹方程.(2)若EF为圆N:x^2+(y-1)^2=1的任一条直线,求PE向
已知坐标平面上一定点C(2,0)和直线l:x=8,P为该平面上一动点,作PQ⊥l,垂足为Q,
且(PC+(1/2)PQ)•(PC-(1/2)PQ)=0.
(1)求动点P的轨迹方程.
(2)若EF为圆N:x²+(y-1)²=1的任一条直线,求PE向量•PF向量的最值.
设P点的坐标为(x,y)
(1).向量PC=(2-x,-y),PQ=(8-x,y-y)=(8-x,0);
故PC+(1/2)PQ=(2-x+(8-x)/2,-y)=(6-(3/2)x,-y);PC-(1/2)PQ=(2-x-(8-x)/2,-y)=(-2-x/2,-y);
(PC+(1/2)PQ)•(PC-(1/2)PQ)=[6-(3/2)x](-2-x/2)+(-y)(-y)=-12+(3/4)x²+y²=0
故得P点的轨迹方程为 x²/16+y²/12=1,即P的轨迹是一个a=4,b=2√3,焦点在x轴上的椭圆.
(2).第二问:【EF为圆N:x²+(y-1)²=1的任一条直线】是什么意思?EF是园的任意弦?请明确
一下,不然不好作.
设pq中点A 那么PC=PA (有三角形中线定理)
设P(x,y)
(x-2)^2+y^2=((x-8)/2)^2
化简得3x^2+4y^2=48