设F为抛物线y2=4X的焦点.A.B.C为该抛物线上三点,若FA+FB+FC=O.则∣FA∣+∣FB∣+∣FC∣=?设A(x1,y1),B(x2,y2),C(x3,y3)抛物线焦点坐标F(1,0),准线方程:x=-1∵FA+FB+FC=O∴点F是△ABC重心则x1+x2+x3=3y1+y2+y3=0而|FA|=x1
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![设F为抛物线y2=4X的焦点.A.B.C为该抛物线上三点,若FA+FB+FC=O.则∣FA∣+∣FB∣+∣FC∣=?设A(x1,y1),B(x2,y2),C(x3,y3)抛物线焦点坐标F(1,0),准线方程:x=-1∵FA+FB+FC=O∴点F是△ABC重心则x1+x2+x3=3y1+y2+y3=0而|FA|=x1](/uploads/image/z/8670357-45-7.jpg?t=%E8%AE%BEF%E4%B8%BA%E6%8A%9B%E7%89%A9%E7%BA%BFy2%3D4X%E7%9A%84%E7%84%A6%E7%82%B9.A.B.C%E4%B8%BA%E8%AF%A5%E6%8A%9B%E7%89%A9%E7%BA%BF%E4%B8%8A%E4%B8%89%E7%82%B9%2C%E8%8B%A5FA%2BFB%2BFC%3DO.%E5%88%99%E2%88%A3FA%E2%88%A3%2B%E2%88%A3FB%E2%88%A3%2B%E2%88%A3FC%E2%88%A3%3D%3F%E8%AE%BEA%28x1%2Cy1%29%2CB%28x2%2Cy2%29%2CC%28x3%2Cy3%29%E6%8A%9B%E7%89%A9%E7%BA%BF%E7%84%A6%E7%82%B9%E5%9D%90%E6%A0%87F%281%2C0%29%2C%E5%87%86%E7%BA%BF%E6%96%B9%E7%A8%8B%EF%BC%9Ax%3D-1%E2%88%B5FA%2BFB%2BFC%3DO%E2%88%B4%E7%82%B9F%E6%98%AF%E2%96%B3ABC%E9%87%8D%E5%BF%83%E5%88%99x1%2Bx2%2Bx3%3D3y1%2By2%2By3%3D0%E8%80%8C%7CFA%7C%3Dx1)
设F为抛物线y2=4X的焦点.A.B.C为该抛物线上三点,若FA+FB+FC=O.则∣FA∣+∣FB∣+∣FC∣=?设A(x1,y1),B(x2,y2),C(x3,y3)抛物线焦点坐标F(1,0),准线方程:x=-1∵FA+FB+FC=O∴点F是△ABC重心则x1+x2+x3=3y1+y2+y3=0而|FA|=x1
设F为抛物线y2=4X的焦点.A.B.C为该抛物线上三点,若FA+FB+FC=O.则∣FA∣+∣FB∣+∣FC∣=?
设A(x1,y1),B(x2,y2),C(x3,y3)
抛物线焦点坐标F(1,0),准线方程:x=-1
∵FA+FB+FC=O
∴点F是△ABC重心
则x1+x2+x3=3
y1+y2+y3=0
而|FA|=x1-(-1)=x1+1
|FB|=x2-(-1)=x2+1
|FC|=x3-(-1)=x3+1
∴|FA|+|FB|+|FC|=x1+1+x2+1+x3+1=(x1+x2+x3)+3=3+3=6
答案是以上,但我有一点不明白
为什么x1+x2+x3=3
设F为抛物线y2=4X的焦点.A.B.C为该抛物线上三点,若FA+FB+FC=O.则∣FA∣+∣FB∣+∣FC∣=?设A(x1,y1),B(x2,y2),C(x3,y3)抛物线焦点坐标F(1,0),准线方程:x=-1∵FA+FB+FC=O∴点F是△ABC重心则x1+x2+x3=3y1+y2+y3=0而|FA|=x1
楼上正解,也可以
FA(向量)=(X1-1,Y1)
FA+FB+FC=(X1-1+X2-1+X3-1,Y1+Y2+Y3)=零向量
则X1-1+X2-1+X3-1=0 即
x1+x2+x3=3
Y1+Y2+Y3=0
重心坐标公式 X=(x1+x2+x3)/3
Y=(y1+y2+y3)/3
由于点F是△ABC重心
所以(x1+x2+x3)/3=1
利用了重心的坐标公式 x1+x2+x3=3x,y1+y2+y3=3y
因为向量FA+FB+FC=O,所以有(X1-1,Y1)+(X2-1,Y2)+(X3-1,Y3)=0,即
X1-1+ X2-1+ X3-1=0,Y1+ Y2+ Y3=0;所以X1+ X2 +X3=3;
重心坐标公式(x1+x2+x3)/3=1 所以x1+x2+x3=3