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    令x=1,得n=(1,-6,3).…9分因为FE=(3,1,0),…10分所以点F到面CDE的距离d=ln·F它=3=346n46…12分/4622.(1)证明:圆C的方程可化为λ(.x2十y2一5)一2x一4y+6=0.…1分x=-1,=4令1-2x-4y十6=0,得v=25、得或…3分2y=5故圆C恒过两个定点,且这两个定点的坐标为(一1,2)和(号,号)…4分(2)解:当λ=1时,圆C的方程可化为x2+y2一2x一4y十1=0.由题知直线1的斜率k存在且不为0,设直线l的方程为y=k(x十1).…5分x2+y2-2x-4y+1=0,联立消去x得(1十2)y2-4k(1+b)y十4k2=0,…7分y=k(x+1),y1十2=4k(1+k)1+21新以8分4k2y12=1+△=16k2(1+k)2-16k(1+2)>0,解得k>0.9分因为1+1=M十业=1+k………………………………………………………10分所以=,解得152…11分又>0,所以=1+⑤2…12分【高二数学·参考答案第5页(共5页)】·24-99B·

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  • 衡水金卷先享题2024答案数学分科综合卷 新教材乙卷A

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