回復 1# Superconan 的帖子
10.
\(\Gamma\)：\(x^2+4xy+8y^2-4=0\)，若\((h,k)\)為\(\Gamma\)上一點，求\(h^2+k^2\)的最小值。
[解答]
\(x^2+4xy+8y^2-4=0\)
\((x+2y)^2+(2y)^2=4\)
故令\(\cases{x+2y=2cos \theta \cr 2y=2sin \theta}\Rightarrow \cases{x=2sin \theta-2 cos \theta \cr y=sin \theta}\)
所求即
\(\displaystyle x^2+y^2=5sin^2 \theta+4cos^2 \theta-8sin\theta cos \theta=cos^2 \theta-8sin \theta cos \theta+4=\frac{1}{2}cos 2\theta-4 sin 2 \theta+\frac{9}{2}\)
又由疊合知\(\displaystyle \frac{9-\sqrt{65}}{2}\le x^2+y^2 \le \frac{9+\sqrt{65}}{2}\)
