每日一题[784]参数方程

已知$A,B$为双曲线$C:\dfrac{x^2}{a^2}-\dfrac{y^2}{b^2}=1$($b>a>0$)上的两点，且以线段$AB$为直径的圆通过坐标原点$O$，则$\triangle AOB$面积的最小值为________．

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正确答案是$\dfrac{a^2b^2}{b^2-a^2}$．

分析与解　不妨设$A\left(r_1\cos\theta,r_1\sin\theta\right)$，$B\left(-r_2\sin\theta,r_2\cos\theta\right)$，则有

$$\begin{cases} \dfrac{r_1^2\cos^2\theta}{a^2}-\dfrac{r_1^2\sin^2\theta}{b^2}=1,\\ \dfrac{r_2^2\sin^2\theta}{a^2}-\dfrac{r_2^2\cos^2\theta}{b^2}=1,\end{cases}$$
于是 $$\dfrac{1}{r_1^2}+\dfrac{1}{r_2^2}=\dfrac{1}{a^2}-\dfrac{1}{b^2},$$
进而可得$\triangle AOB$的面积 $$S=\dfrac 12r_1r_2=\dfrac 12r_1r_2\cdot \left(\dfrac{1}{r_1^2}+\dfrac{1}{r_2^2}\right)\cdot \dfrac{1}{\dfrac{1}{a^2}-\dfrac{1}{b^2}}=\dfrac 12\left(\dfrac{r_2}{r_1}+\dfrac{r_1}{r_2}\right)\cdot \dfrac{a^2b^2}{b^2-a^2}\geqslant \dfrac{a^2b^2}{b^2-a^2},$$ 等号当$r_1=r_2$时取得．因此所求面积的最小值为$\dfrac{a^2b^2}{b^2-a^2}$．