第(3)道数学题怎么做?
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则直线是y-1=k(x-0)
y=kx+1
代入3x²-(k²x²+2kx+1)=1
(3-k²)x²-2kx-2=0
则x1+x2=2k/(3-k²),x1x2=-2/(3-k²)
O在圆上,而A和B是直径的两个端点
所以OA垂直OB
则k1k2=-1
即y1/x1*y2/x2=-1
x1x2+y1y2=0
x1x2+(kx1+1)(kx2+1)=0
(1+k²)x1x2+k(x1+x2)+1=0
则-2(1+k²)/(3-k²)+2k²/(3-k²)+1=0
-2(1+k²)+2k²+(3-k²)=0
-2+3-k²=0
k=±1
所以是y=±x+1
则直线是y-1=k(x-0)
y=kx+1
代入3x²-(k²x²+2kx+1)=1
(3-k²)x²-2kx-2=0
则x1+x2=2k/(3-k²),x1x2=-2/(3-k²)
O在圆上,而A和B是直径的两个端点
所以OA垂直OB
则k1k2=-1
即y1/x1*y2/x2=-1
x1x2+y1y2=0
x1x2+(kx1+1)(kx2+1)=0
(1+k²)x1x2+k(x1+x2)+1=0
则-2(1+k²)/(3-k²)+2k²/(3-k²)+1=0
-2(1+k²)+2k²+(3-k²)=0
-2+3-k²=0
k=±1
所以是y=±x+1
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