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計算第2題
這題其實不難，就計算量大了點，考試時一定要先跳過
設直線AB的方程式為\(y=mx+k\)，\(A\left( {{x}_{1}},m{{x}_{1}}+k \right),B\left( {{x}_{2}},m{{x}_{2}}+k \right)\)
\(\begin{align}
& \frac{{{x}^{2}}}{25}+\frac{{{\left( mx+k \right)}^{2}}}{16}=1 \\
& \left( 25{{m}^{2}}+16 \right){{x}^{2}}+50mkx+\left( 25{{k}^{2}}-400 \right)=0 \\
& {{\left( 50mk \right)}^{2}}-4\left( 25{{m}^{2}}+16 \right)\left( 25{{k}^{2}}-400 \right)=0 \\
& {{k}^{2}}=25{{m}^{2}}+16 \\
& {{x}_{1}}=-\frac{50mk}{2\left( 25{{m}^{2}}+16 \right)}=-\frac{50mk}{2{{k}^{2}}}=-\frac{25m}{k} \\
& \\
& {{x}^{2}}+{{\left( mx+k \right)}^{2}}={{R}^{2}} \\
& \left( {{m}^{2}}+1 \right){{x}^{2}}+2mkx+\left( {{k}^{2}}-{{R}^{2}} \right)=0 \\
& {{\left( 2mk \right)}^{2}}-4\left( {{m}^{2}}+1 \right)\left( {{k}^{2}}-{{R}^{2}} \right)=0 \\
& {{k}^{2}}={{R}^{2}}\left( {{m}^{2}}+1 \right) \\
& {{x}_{2}}=-\frac{2mk}{2\left( {{m}^{2}}+1 \right)}==-\frac{m{{R}^{2}}}{k} \\
& \\
& {{k}^{2}}={{R}^{2}}\left( {{m}^{2}}+1 \right)=25{{m}^{2}}+16 \\
& {{m}^{2}}=\frac{16-{{R}^{2}}}{{{R}^{2}}-25} \\
& \\
& {{\overline{AB}}^{2}}={{\left( {{x}_{1}}-{{x}_{2}} \right)}^{2}}+{{\left( m{{x}_{1}}-m{{x}_{2}} \right)}^{2}}=\left( {{m}^{2}}+1 \right){{\left( {{x}_{1}}-{{x}_{2}} \right)}^{2}} \\
& =\left( {{m}^{2}}+1 \right){{\left( -\frac{25m}{k}+\frac{m{{R}^{2}}}{k} \right)}^{2}} \\
& =\left( {{m}^{2}}+1 \right)\times \frac{\frac{16-{{R}^{2}}}{{{R}^{2}}-25}{{\left( {{R}^{2}}-25 \right)}^{2}}}{{{R}^{2}}\left( {{m}^{2}}+1 \right)} \\
& =\frac{\left( 16-{{R}^{2}} \right)\left( {{R}^{2}}-25 \right)}{{{R}^{2}}} \\
& =41-\left( {{R}^{2}}+\frac{400}{{{R}^{2}}} \right) \\
& \le 41-40 \\
& =1 \\
\end{align}\)
\(\overline{AB}\)長的最大值是1，等號成立於\(R=2\sqrt{5}\)時
