Question

19题答案？

Answer 1

设圆心(x₁,y₁)

(x-x₁)²+(y-y₁)²=R² 过定点(0,1) 代入：

(0-x₁)²+(1-y₁)²=R²→R²=x₁²+(1-y₁)²

与y=-1相切：圆心到直线的距离=R

(y₁+1)²=x₁²+(1-y₁)²

4y₁=x₁²→C：y=¼x²

(2) 直线y=kx+1

kx+1=¼x²→x²-4kx-4=0

x₁+x₂=4k x₁·x₂=-4

(x₁-x₂)²=(x₁+x₂)²-4x₁·x₂=16k²+16

(y₁-y₂)²=(kx₁+1-kx₂-1)²=k²(x₁-x₂)²

(x₁-x₂)²+(y₁-y₂)²=8²

16(1+k²)²=64→k=±1

x²-4x-4=0→x₁=½(4±4√2)=2±2√2

A(2-2√2,3-2√2) B(2+2√2,3+2√2)

QA⊥QB

设Q(a,-1)

[(4-2√2)/(2-2√2-a)]·[(4+2√2)/(2+2√2-a)]=-1

a=2→Q(-2,1)

Q到AB(x-y+1=0)的距离：|-2-1+1|/√2=√2

S=½·8·√2=4√2  (k=-1,结果一样）