Question

题目第一问怎么做，解答看不懂 求第一问详细过程

Answer 1

ln: y=x-√(2n) (1)
Cn : x^2+y^2=2an+n+2 (2)
a1=1
a(n+1) = (1/4)|AnBn|^2
let
An=(x1,y1) ,Bn=(x2,y2)
sub (1) into (2)
x^2+y^2=2an+n+2
x^2 +(x-√(2n))^2 = 2an+n+2
2x^2 - 2√(2n)x -(2an - n +2) = 0
x1+x2 =√(2n)
x1.x2 = -(2an - n +2)/2
(x1-x2)^2
=( x1+x2)^2 - 4x1.x2
=2n + 2(2an - n +2)
= 4an + 4
Similarly
x^2+y^2=2an+n+2
(y+√(2n))^2 +y^2 = 2an+n+2
2y^2 +2√(2n)y -(2an-n+2) =0
y1+y2 =-√(2n)
y1.y2 = -(2an-n+2)/2
(y1-y2)^2
=(y1+y2)^2 -4y1y2
=2n +2(2an-n+2)
=4an +4
|AnBn| ^2
=(x1-x2)^2+(y1-y2)^2
=8an +8
a(n+1) = (1/4)|AnBn|^2
a(n+1) = 2an +2
a(n+1) +2 = 2(an + 2)
=> {an + 2} 是等比数列, q=2
an + 2 = 2^(n-1) .(a1 + 2)
= 3. 2^(n-1)
an = -2 +3. 2^(n-1)