回復 1# jackyxul4 的帖子
P 的橫坐標應大於 2
下面式子的直線PB和直線PC，符號沒有出來
\(\begin{align}
& P\left( x{}_{0},{{y}_{0}} \right),B\left( 0,b \right),C\left( 0,c \right),b>0>c,{{x}_{0}}>2 \\
& \\
& :\left( {{y}_{0}}-b \right)x-{{x}_{0}}y+{{x}_{0}}b=0 \\
& :\left( {{y}_{0}}-c \right)x-{{x}_{0}}y+{{x}_{0}}c=0 \\
& \\
& \frac{\left| {{y}_{0}}-b+{{x}_{0}}b \right|}{\sqrt{{{\left( {{y}_{0}}-b \right)}^{2}}+{{\left( -{{x}_{0}} \right)}^{2}}}}=\frac{\left| {{y}_{0}}-c+{{x}_{0}}c \right|}{\sqrt{{{\left( {{y}_{0}}-c \right)}^{2}}+{{\left( -{{x}_{0}} \right)}^{2}}}}=1 \\
& \left( {{x}_{0}}-2 \right){{b}^{2}}+2{{y}_{0}}b-{{x}_{0}}=\left( {{x}_{0}}-2 \right){{c}^{2}}+2{{y}_{0}}c-{{x}_{0}}=0 \\
\end{align}\)
b和c是\(\left( {{x}_{0}}-2 \right){{t}^{2}}+2{{y}_{0}}t-{{x}_{0}}=0\)的二根
\(\begin{align}
& \overline{BC}=b-c=\frac{2\sqrt{{{\left( 2{{y}_{0}} \right)}^{2}}+4{{x}_{0}}\left( {{x}_{0}}-2 \right)}}{2\left( {{x}_{0}}-2 \right)}=\frac{2{{x}_{0}}}{{{x}_{0}}-2} \\
& \Delta PBC=\frac{1}{2}\times \frac{2{{x}_{0}}}{{{x}_{0}}-2}\times {{x}_{0}}={{x}_{0}}-2+\frac{4}{{{x}_{0}}-2}+4\ge 4+4=8 \\
\end{align}\)
[ 本帖最後由 thepiano 於 2016-4-21 12:52 PM 編輯 ]
