在直角坐标系中，$O$ 是原点，$A$、$B$ 是第一象限内的点，并且 $A$ 在直线 $y = \tan \theta \cdot x$ 上（其中 $\theta \in \left( {\dfrac{{{\pi }}}{4} , \dfrac{{{\pi }}}{2}} \right)$），$\left| {OA} \right| = \dfrac{1}{{\sqrt 2 - \cos \theta }}$，$B$ 是双曲线 ${x^2} - {y^2} = 1$ 上使 $\triangle OAB$ 的面积最小的点，求：当 $\theta $ 在 $\left( {\dfrac{{{\pi }}}{4} , \dfrac{{{\pi }}}{2}} \right)$ 中取什么值时，$\triangle OAB$ 的面积最大，最大值是多少？
【难度】
【出处】
2001年复旦大学保送生招生测试
【标注】
【答案】
当 $\theta = \arccos \dfrac{{\sqrt 2 }}{4}$ 时，$\triangle OAB$ 的面积取得最大值为 $\dfrac{{\sqrt 6 }}{6}$
【解析】
根据题意$$\begin{cases}
{x_0}^2 - {y_0}^2 = 1 ,\hfill \cr
\dfrac{{{x_0}}}{{{y_0}}} = \tan \theta .\hfill \cr
\end{cases}$$所以 ${x_0} = \dfrac{{\sin \theta }}{{\sqrt {{{\sin }^2}\theta - {{\cos }^2}\theta } }}$，${y_0} = \dfrac{{\cos \theta }}{{\sqrt {{{\sin }^2}\theta - {{\cos }^2}\theta } }}$．故\[\begin{split}d\left( {B , OA} \right) &= \dfrac{{{x_0}\tan \theta - {y_0}}}{{\sqrt {{{\tan }^2}\theta + 1} }}\\& = {x_0}\sin \theta - {y_0}\cos \theta \\& = \sqrt {{{\sin }^2}\theta - {{\cos }^2}\theta } .\end{split}\]所以$${S_{\triangle OAB}}^2 = \dfrac{{{{\sin }^2}\theta - {{\cos }^2}\theta }}{{4{{\left( {\sqrt 2 - \cos \theta } \right)}^2}}} = \dfrac{{1 - 2{{\cos }^2}\theta }}{{4{{\left( {\sqrt 2 - \cos \theta } \right)}^2}}}.$$令 $t = \sqrt 2 - \cos \theta $，则 $\cos \theta = \sqrt 2 - t$，所以\[\begin{split}S_{\triangle OAB}^2 &= \dfrac{{1 - 2{{\left( {\sqrt 2 - t} \right)}^2}}}{{4{t^2}}} \\&= \dfrac{1}{4} \cdot \dfrac{{ - 2{t^2} + 4\sqrt 2 t - 3}}{{{t^2}}}\\& = \dfrac{1}{4}\left[ { - 3 \cdot {{\left( {\dfrac{1}{t}} \right)}^2} + 4\sqrt 2 \cdot \dfrac{1}{t} - 2} \right]\\& = \dfrac{1}{4}\left[ { - 3{{\left( {\dfrac{1}{t} - \dfrac{{2\sqrt 2 }}{3}} \right)}^2} + \dfrac{2}{3}} \right] \\& \leqslant \dfrac{1}{6}.\end{split}\]（当且仅当 $t = \dfrac{{3\sqrt 2 }}{4}$，即 $\cos \theta = \dfrac{{\sqrt 2 }}{4}$ 时取得等号）．
因此当 $\theta = \arccos \dfrac{{\sqrt 2 }}{4}$ 时，$\triangle OAB$ 的面积取得最大值为 $\dfrac{{\sqrt 6 }}{6}$．
{x_0}^2 - {y_0}^2 = 1 ,\hfill \cr
\dfrac{{{x_0}}}{{{y_0}}} = \tan \theta .\hfill \cr
\end{cases}$$所以 ${x_0} = \dfrac{{\sin \theta }}{{\sqrt {{{\sin }^2}\theta - {{\cos }^2}\theta } }}$，${y_0} = \dfrac{{\cos \theta }}{{\sqrt {{{\sin }^2}\theta - {{\cos }^2}\theta } }}$．故\[\begin{split}d\left( {B , OA} \right) &= \dfrac{{{x_0}\tan \theta - {y_0}}}{{\sqrt {{{\tan }^2}\theta + 1} }}\\& = {x_0}\sin \theta - {y_0}\cos \theta \\& = \sqrt {{{\sin }^2}\theta - {{\cos }^2}\theta } .\end{split}\]所以$${S_{\triangle OAB}}^2 = \dfrac{{{{\sin }^2}\theta - {{\cos }^2}\theta }}{{4{{\left( {\sqrt 2 - \cos \theta } \right)}^2}}} = \dfrac{{1 - 2{{\cos }^2}\theta }}{{4{{\left( {\sqrt 2 - \cos \theta } \right)}^2}}}.$$令 $t = \sqrt 2 - \cos \theta $，则 $\cos \theta = \sqrt 2 - t$，所以\[\begin{split}S_{\triangle OAB}^2 &= \dfrac{{1 - 2{{\left( {\sqrt 2 - t} \right)}^2}}}{{4{t^2}}} \\&= \dfrac{1}{4} \cdot \dfrac{{ - 2{t^2} + 4\sqrt 2 t - 3}}{{{t^2}}}\\& = \dfrac{1}{4}\left[ { - 3 \cdot {{\left( {\dfrac{1}{t}} \right)}^2} + 4\sqrt 2 \cdot \dfrac{1}{t} - 2} \right]\\& = \dfrac{1}{4}\left[ { - 3{{\left( {\dfrac{1}{t} - \dfrac{{2\sqrt 2 }}{3}} \right)}^2} + \dfrac{2}{3}} \right] \\& \leqslant \dfrac{1}{6}.\end{split}\]（当且仅当 $t = \dfrac{{3\sqrt 2 }}{4}$，即 $\cos \theta = \dfrac{{\sqrt 2 }}{4}$ 时取得等号）．
因此当 $\theta = \arccos \dfrac{{\sqrt 2 }}{4}$ 时，$\triangle OAB$ 的面积取得最大值为 $\dfrac{{\sqrt 6 }}{6}$．
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