已知\(\displaystyle A(a,b),B(-a,b),C(0,\frac{1}{2})\)為橢圓\(\Gamma\):\(x^2+4y^2=1\)上的三點,若過\(A,B,C\)三點的圓半徑為\(r\),則\(\displaystyle \lim_{a\to 0}r=\) 。
補第二題的想法
首先知a^2+4b^2=1,
因AC斜率為(b-1/2)/a=(2b-1)/(2a), AC中點(a/2, (b+1/2)/2)
=> AC中垂線為y-(b+1/2)/2=[(-2a)/(2b-1)](x-a/2)
代x=0(AB中垂線), 解得圓心坐標( 0, (4a^2+4b^2-1)/(8b-4) )=( 0, (3-12b^2)/(8b-4))
因a->0 時, b->+-1/2
lim(b->-1/2) [(3-12b^2)/(8b-4)]=0
lim(b->1/2) [(3-12b^2)/(8b-4)]=lim(b->1/2) [(-24b)/8]=-3/2 (L'hospital's Rule)
故lim(a->0) (r) =lim(b->+-1/2) (r)
=1/2-0 或 1/2-(-3/2)
=1/2 或 2