回復 1# swallow7103 的帖子
計算第 2 題
(1)
\(\begin{align}
& 直線 AB:y=mx+n \\
& {{\left( mx+n \right)}^{2}}=6x \\
& {{m}^{2}}{{x}^{2}}+\left( 2mn-6 \right)x+{{n}^{2}}=0 \\
& {{x}_{1}}+{{x}_{2}}=-\frac{2mn-6}{{{m}^{2}}}=4 \\
& n=\frac{3}{m}-2m \\
& y=mx+\frac{3}{m}-2m \\
\end{align}\)
\(\overline{AB}\)中點為\(M\left( 2,\frac{3}{m} \right)\)
垂直平分線：\(y-\frac{3}{m}=-\frac{1}{m}\left( x-2 \right)\)交\(x\)軸於\(C\left( 5,0 \right)\)
(2)
\(\begin{align}
& {{y}_{0}}=\frac{3}{m} \\
& {{\left( 2mn-6 \right)}^{2}}-4{{m}^{2}}{{n}^{2}}>0 \\
& mn<\frac{3}{2} \\
& m\left( \frac{3}{m}-2m \right)<\frac{3}{2} \\
& m>\frac{\sqrt{3}}{2},m<\text{-}\frac{\sqrt{3}}{2} \\
& -2\sqrt{3}<{{y}_{0}}<2\sqrt{3} \\
\end{align}\)
(3)
直線\(AB\)交\(x\)軸於\(D\left( -\frac{n}{m},0 \right)\)
\(\begin{align}
& \left\{ \begin{align}
& {{y}^{2}}=6x \\
& y=mx+n \\
\end{align} \right. \\
& {{y}^{2}}=6\left( \frac{y-n}{m} \right) \\
& m{{y}^{2}}-6y+6n=0 \\
& {{y}_{1}}+{{y}_{2}}=\frac{6}{m},{{y}_{1}}{{y}_{2}}=\frac{6n}{m} \\
& \Delta ABC=\frac{1}{2}\left| 5+\frac{n}{m} \right|\left| {{y}_{1}}-{{y}_{2}} \right|=\frac{1}{2}\left| 5+\frac{n}{m} \right|\sqrt{{{\left( {{y}_{1}}+{{y}_{2}} \right)}^{2}}-4{{y}_{1}}{{y}_{2}}} \\
& =3\left( 1+\frac{1}{{{m}^{2}}} \right)\sqrt{12-\frac{9}{{{m}^{2}}}} \\
& \\
\end{align}\)
令
\(\begin{align}
& t=\sqrt{12-\frac{9}{{{m}^{2}}}} \\
& \frac{1}{{{m}^{2}}}=\frac{4}{3}-\frac{{{t}^{2}}}{9} \\
& \Delta ABC=3\left( 1+\frac{4}{3}-\frac{{{t}^{2}}}{9} \right)t=-\frac{{{t}^{3}}}{3}+7t \\
\end{align}\)
最大值為\(\frac{14}{3}\sqrt{7}\)
[ 本帖最後由 thepiano 於 2022-4-16 16:17 編輯 ]
