由动点P引圆x2+y2=10的两条切线PA,PB,直线PA,PB的斜率分别为k1,k2. (1)若k1+k2+k1k2=

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  • 1) 若k1+k2+k1×k2=-1,求动点P的轨迹方程

    设点P为(a,b),

    直线为y-b=k(x-a)

    代入圆方程

    x²+(kx-ak+b)²=10

    (1+k²)x²-2kx(ak-b)+(ak-b)²-10=0

    因直线与圆相切则方程仅有一实根

    则4k²(ak-b)²=4(1+k²)[(ak-b)²-10]

    a²k^4-2abk³+b²k²=a²k^4-2k³ab+k²(b²-10)+a²k²-2abk+b²-10

    (a²-10)k²-2abk+b²-10=0

    则k1+k2=2ab/(a²-10),k1*k2=(b²-10)/(a²-10)

    因k1+k2+k1×k2=-1,

    则2ab/(a²-10)+(b²-10)/(a²-10)=-1

    2ab+a²-10+b²-10=0

    (a+b)²=20

    P点轨迹为x+y=±2√5两直线,除点(±√5,±√5)两个点以为.

    2) 若点P在直线x+y=m上,且AP⊥BP,求实数m的取值范围

    已证k1*k2=(b²-10)/(a²-10)

    AP⊥BP

    则k1*k2=-1

    则(b²-10)/(a²-10)=-1

    a²+b²=20

    P点轨迹为x²+y²=20

    有P在直线x+y=m上

    则(m-y)²+y²-20=0

    y²-my+m²/2-10=0

    则m²>=4(m²/2-10)