回復 1# gonm 的帖子
2. 昨天有人問，順便來 po 了一下
令 \( P(2\alpha^{2},2\alpha) \), 切線 \( y=mx+2\alpha-2\alpha^{2}m \). 代入圓得
\( x^{2}-2x+1+(m^{2}x^{2}+4(\alpha-\alpha^{2}m)mx+4(\alpha-\alpha^{2}m)^{2})=1 \)
判別式為 0
\( \begin{aligned} & (m^{2}+1)x^{2}+(4\alpha m-4\alpha^{2}m^{2}-2)x+4\alpha^{2}(1-\alpha m)^{2}=0\\
\Rightarrow & (4\alpha m-4\alpha^{2}m^{2}-2)^{2}-16(m^{2}+1)\alpha^{2}(1-\alpha m)^{2}=0\\
\Rightarrow & 4(\alpha^{2}-\alpha^{4})m^{2}+(8\alpha^{3}-4\alpha)m+1-4\alpha^{2}=0\\
\Rightarrow & \Delta m=\frac{\sqrt{16\alpha^{2}(2\alpha^{2}-1)^{2}-16(\alpha^{2}-\alpha^{4})(1-4\alpha^{2})}}{4(\alpha^{2}-\alpha^{4})}=\frac{1}{1-\alpha^{2}}.\end{aligned} \)
\( x=0 \) 代入直線，得 \( \overline{BC}=\left|\frac{2\alpha^{2}}{1-\alpha^{2}}\right|\)
畫圖，知僅 \( \alpha>1 \) 時，是內切圓，此時三角形面積 \( =\frac{1}{2}\cdot(2\alpha^{2})\cdot\frac{2\alpha^{2}}{1-\alpha^{2}}=2\cdot\frac{\alpha^{4}}{\alpha^{2}-1} =2\cdot(2+\alpha^{2}-1+\frac{1}{\alpha^{2}-1})\geq8 \) by 算幾。
不過以上太暴力，應該會有其它好的方法吧
