已知直线l:x-ny=0(n∈N*),圆M:(x+1) 2 +(y+1) 2 =1,抛物线φ:y=(x-1) 2 ,又l

已知直线l:x-ny=0(n∈N*),圆M:(x+1) 2 +(y+1) 2 =1,抛物线φ:y=(x-1) 2 ,又l与M交于点A、B,l与φ交于点C、D,求
lim
n→∞
|AB | 2
|CD | 2
fhbrd3 1年前 已收到1个回答 举报

whenit 幼苗

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设圆心M(-1,-1)到直线l的距离为d,则d 2 =
(n-1) 2

n 2 +1 .
又r=1,∴|AB| 2 =4(1-d 2 )=
8n
1+
n 2 .
设点C(x 1 ,y 1 ),D(x 2 ,y 2 ),


x-ny=0
y=(x-1 ) 2 ⇒nx 2 -(2n+1)x+n=0,
∴x 1 +x 2 =,x 1 •x 2 =1.
∵(x 1 -x 2 2 =(x 1 +x 2 2 -4x 1 x 2 =
4n+1

n 2 ,(y 1 -y 2 2 =(
x 1
n -
x 2
n ) 2 =
4n+1
n 4 ,
∴|CD| 2 =(x 1 -x 2 2 +(y 1 -y 2 2
=
1
n 4 (4n+1)(n 2 +1).

lim
n→∞
|A B | 2
|CD | 2 =
lim
n→∞
8 n 5
(4n+1) ( n 2 +1) 2 =
lim
n→∞
8
(4+
1
n ) (1+
1
n ) 2 =2.

1年前

5
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