Question

由动点p引圆x^2+y^2=10的俩条切线PA PB 的斜率分别是

由动点p引圆x^2+y^2=10的俩条切线PA PB 的斜率分别是k1 k2 1 若k1+k2+k1k2=-1 求p点的轨迹方程2若在直线x+y=m上且PA⊥PB 求实数m的取值范围

Answer 1

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
x�0�5+(kx-ak+b)�0�5=10
(1+k�0�5)x�0�5-2kx(ak-b)+(ak-b)�0�5-10=0
因直线与圆相切则方程仅有一实根
则4k�0�5(ak-b)�0�5=4(1+k�0�5)[(ak-b)�0�5-10]
<=>a�0�5k^4-2abk�0�6+b�0�5k�0�5=a�0�5k^4-2k�0�6ab+k�0�5(b�0�5-10)+a�0�5k�0�5-2abk+b�0�5-10
<=>(a�0�5-10)k�0�5-2abk+b�0�5-10=0
则k1+k2=2ab/(a�0�5-10),k1*k2=(b�0�5-10)/(a�0�5-10)
因k1+k2+k1×k2=-1,
则2ab/(a�0�5-10)+(b�0�5-10)/(a�0�5-10)=-1
2ab+a�0�5-10+b�0�5-10=0
(a+b)�0�5=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=(b�0�5-10)/(a�0�5-10)
AP⊥BP
则k1*k2=-1
则(b�0�5-10)/(a�0�5-10)=-1
a�0�5+b�0�5=20
P点轨迹为x�0�5+y�0�5=20
有P在直线x+y=m上
则(m-y)�0�5+y�0�5-20=0
y�0�5-my+m�0�5/2-10=0
则m�0�5>=4(m�0�5/2-10)
m�0�5<=40
则-2√10≤m≤2√10