设L1：y=k(x-c),其中c=√（a^2-b^2),
代入椭圆方程得b^2x^2+a^2k^2(x-c)^2=a^2b^2,
(b^2+a^2k^2)x^2-2a^2ck^2x+a^2(c^2k^2-b^2)=0,
设A(x1,y1),B(x2,y2),则
x1+x2=2a^2ck^2/(b^2+a^2k^2),
x1x2=a^2(c^2k^2-b^2)/(b^2+a^2k^2),
|x1-x2|=√[(x1+x2)^2-4x1x2]
=2ab^2√（1+k^2)/(b^2+a^2k^2),
|FA|=|x1-c|√（1+k^2),
|FB|=|x2-c|√（1+k^2),
∴|FA|*|FB|=|(x1-c)(x2-c)|(1+k^2)
=|[x1x2-c(x1+x2)+c^2]|(1+k^2)
=|[a^2(c^2k^2-b^2)-2a^2c^2k^2+c^2(b^2+a^2k^2)]|(1+k^2)/(b^2+a^2k^2)
=b^4(1+k^2)/(b^2+a^2k^2),
FC*FD=b^4(1+1/k^2)/(b^2+a^2/k^2)=b^4(1+k^2)/(a^2+b^2k^2)≠FA*FB,
(1)1/FA + 1/FB
=(FA+FB)/(FA*FB)
=(x1-x2)√（1+k^2)/(-|FA|*|FB|)
=2a/[b^2*√（1+k^2）],
∵L1⊥L2,
∴以-1/k代k,得
1/FC + 1/FD
=2a/[b^2*√(1+1/k^2)]
≠1/FA + 1/FB,
请检查题目.