已知圆C:x²+(y-1)²=5,直线L:mx-y+1-m=0,设该直线与圆相交于AB两点,若|AB|=√17,求L的倾斜
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mx-y+1-m=0 =>y=mx+1-m
代入圆方程 =>
x²+(mx-m)²=5 =>
(1+m²)x²-2m²x+m²-5=0 (1)
设两交点为(x1,y1)(x2,y2)
|AB|=根号[(x2-x1)^2+(y2-y1)^2]=根号(1+m²)|x2-x1| (x2,x1为方程1的两个根)
|x2-x1|=根号[(x1+x2)^2-4x1x2]=根号{2m²/(1+m²)]^2-4(m²-5)/(1+m²)}=√17/根号(1+m²)
=>(2m²)^2-4(m²-5)(1+m²)=17(1+m²)
=>3-m²=0 m=根号3,负根号3.(即为斜率)
|AB|=根号[4m^4/(1+m²)-4(m²-5)]=根号[(20+16m²)/(1+m^2)]=根号[16+4/(1+m^2)]
=> m=0 |AB|=20 取最大值
m=无穷大时,|AB|=16 取最小值
代入圆方程 =>
x²+(mx-m)²=5 =>
(1+m²)x²-2m²x+m²-5=0 (1)
设两交点为(x1,y1)(x2,y2)
|AB|=根号[(x2-x1)^2+(y2-y1)^2]=根号(1+m²)|x2-x1| (x2,x1为方程1的两个根)
|x2-x1|=根号[(x1+x2)^2-4x1x2]=根号{2m²/(1+m²)]^2-4(m²-5)/(1+m²)}=√17/根号(1+m²)
=>(2m²)^2-4(m²-5)(1+m²)=17(1+m²)
=>3-m²=0 m=根号3,负根号3.(即为斜率)
|AB|=根号[4m^4/(1+m²)-4(m²-5)]=根号[(20+16m²)/(1+m^2)]=根号[16+4/(1+m^2)]
=> m=0 |AB|=20 取最大值
m=无穷大时,|AB|=16 取最小值
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