1) 若k1+k2+k1×k2=-1,求動點P的軌跡方程
設點P為(a,b),
直線為y-b=k(x-a)
代入圓方程
x²+(kx-ak+b)²=10
(1+k²)x²-2kx(ak-b)+(ak-b)²-10=0
因直線與圓相切則方程僅有一實根
則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
則k1+k2=2ab/(a²-10),k1*k2=(b²-10)/(a²-10)
因k1+k2+k1×k2=-1,
則2ab/(a²-10)+(b²-10)/(a²-10)=-1
2ab+a²-10+b²-10=0
(a+b)²=20
P點軌跡為x+y=±2√5兩直線,除點(±√5,±√5)兩個點以為.
2) 若點P在直線x+y=m上,且AP⊥BP,求實數(shù)m的取值范圍
已證k1*k2=(b²-10)/(a²-10)
AP⊥BP
則k1*k2=-1
則(b²-10)/(a²-10)=-1
a²+b²=20
P點軌跡為x²+y²=20
有P在直線x+y=m上
則(m-y)²+y²-20=0
y²-my+m²/2-10=0
則m²>=4(m²/2-10)
m²