M(-3,0)N(3,0)B(1,0) 动圆C与MN切于B 过M ,N与C相切的两直线交于P 求P轨迹

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  • 由已知,设PM,PN分别与圆C相切于R、Q,根据圆的切线长定理,有PQ=PR,MQ=MB,NR=NB;

    所以

    PM-PN=QM-RN=MB-NB=21)

    设等差数列{an}公差为d.

    a6-a2=4d=21-5=16 d=4

    a1=a2-d=5-4=1

    an=a1+(n-1)d=1+4(n-1)=4n-3

    1/an=1/(4n-3)

    S(2n+1)-Sn=a1+a2+...+an+a(n+1)+...+a(2n+1)-(a1+a2+...+an)

    =a(n+1)+a(n+2)+...+a(2n+1)

    a[(n+1)+1]+a[(n+1)+2]+...+a[2(n+1)+1]-[a(n+1)+a(n+2)+...+a(2n+1)]

    =a(n+2)+a(n+3)+...+a(2n+1)+a(2n+2)+a(2n+3)-[a(n+1)+a(n+2)+...+a(2n+1)]

    =a(2n+2)+a(2n+3)-a(n+1)

    =1/[4(2n+2) -3]+1/[4(2n+3) -3] -1/[4(n+1)-3]

    =1/(8n+5)+1/(8n+9)-2/(8n+2)