已知 $O$ 是坐标原点，$A,B$ 为椭圆 $C:\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2} = 1$ 上的任意两点．设 $| OA|=r_1,|OB|=r_2$，点 $O$ 到直线 $AB$ 的距离为 $d$，求证：$OA\perp OB$ 的充分必要条件是 $\dfrac{1}{d^2}=\dfrac{1}{r_1^2}+\dfrac{1}{r_2^2}=\dfrac{1}{a^2}+\dfrac{1}{b^2}$．
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【解析】
{\bf{方法一}}
若 $OA\perp OB$，设点 $A$ 坐标为 $\left(r_1\cos \theta,r_1 \sin \theta \right)$，则点 $B$ 坐标为 $\left(r_2\cos \left(\theta \pm \dfrac{\pi}{2}\right),r_2 \sin \left(\theta \pm \dfrac{\pi}{2}\right) \right)$．
将 $A,B$ 坐标代入椭圆方程，可得$$\begin{cases}\dfrac{1}{r_1^2}&=\dfrac{\cos ^2 \theta}{a^2}+\dfrac{\sin ^2 \theta}{b^2},\\ \dfrac{1}{r_2^2}&=\dfrac{\sin ^2 \theta}{a^2}+\dfrac{\cos ^2 \theta}{b^2}. \end{cases}$$以上两式相加，即得 $\dfrac{1}{r_1^2}+\dfrac{1}{r_2^2}=\dfrac{1}{a^2}+\dfrac{1}{b^2}$．又易证 $\dfrac{1}{d^2}=\dfrac{1}{r_1^2}+\dfrac{1}{r_2^2}$，故$$\dfrac{1}{d^2}=\dfrac{1}{r_1^2}+\dfrac{1}{r_2^2}=\dfrac{1}{a^2}+\dfrac{1}{b^2}.$$反之，若 $\dfrac{1}{d^2}=\dfrac{1}{r_1^2}+\dfrac{1}{r_2^2}=\dfrac{1}{a^2}+\dfrac{1}{b^2}$，则类似可证 $OA\perp OB$．
证毕．
{\bf{方法二}}
设点 $A$ 坐标为 $\left(a\cos \alpha,b \sin \alpha \right)$，点 $B$ 坐标为 $\left(a\cos \beta,b \sin \beta \right)$．因为 $OA\perp OB$，所以$$a^2\cos{\alpha}\cos{\beta}+b^2\sin{\alpha}\sin{\beta}=0.$$因为\[\begin{split}a^2\cos^2{\alpha}&=a^2\cos^2{\alpha}\sin^2{\beta}+a^2\cos^2{\alpha}\cos^2{\beta},\\b^2\sin^2{\alpha}&=b^2\sin^2{\alpha}\sin^2{\beta}+b^2\sin^2{\alpha}\cos^2{\beta},\\a^2\cos^2{\beta}&=a^2\cos^2{\beta}\sin^2{\alpha}+a^2\cos^2{\beta}\cos^2{\alpha},\\b^2\sin^2{\beta}&=b^2\sin^2{\beta}\sin^2{\alpha}+b^2\sin^2{\beta}\cos^2{\alpha},\end{split}\]所以\[\begin{split}&a^2\cos^2{\alpha}+b^2\sin^2{\alpha}+a^2\cos^2{\beta}+b^2\sin^2{\beta}\\=& \left(a^2+b^2\right)\left(\sin^2{\alpha}\cos^2{\beta}+\cos^2{\alpha}\sin^2{\beta}\right)+2a^2\cos^2{\alpha}\cos^2{\beta}+2b^2\sin^2{\alpha}\sin^2{\beta}\\=&\left(a^2+b^2\right)\left(\sin^2{\alpha}\cos^2{\beta}+\cos^2{\alpha}\sin^2{\beta}\right)-2b^2\sin{\alpha}\sin{\beta}cos{\alpha}\cos{\beta}-2a^2\sin{\alpha}\sin{\beta}cos{\alpha}\cos{\beta}\\=&\left(a^2+b^2\right)\left(\sin{\alpha}\cos{\beta}-\cos{\alpha}\sin{\beta}\right)^2\\=&\left(a^2+b^2\right)\sin^2{\left(\alpha-\beta\right) }. \end{split}\]因为\[a^4\cos^2{\alpha}\cos^2{\beta}+b^4\sin^2{\alpha}\sin^2{\beta}=-2a^2b^2\sin{\alpha}\cos{\beta}\cos{\alpha}\sin{\beta},\]所以\[\begin{split}&a^4\cos^2{\alpha}\cos^2{\beta}+b^4\sin^2{\alpha}\sin^2{\beta}+a^2b^2\left(\sin^2{\alpha}\cos^2{\beta}+\cos^2{\alpha}\sin^2{\beta}\right)\\=&a^2b^2\sin^2{\left(\alpha-\beta\right)}. \end{split}\]故\[\begin{split}\dfrac{1}{r_1^2}+\dfrac{1}{r_2^2}&=\dfrac{1}{a^2\cos^2{\alpha}+b^2\sin^2{\alpha}}+\dfrac{1}{a^2\cos^2{\beta}+b^2\sin^2{\beta}}\\&= \dfrac{a^2\cos^2{\alpha}+b^2\sin^2{\alpha}+a^2\cos^2{\beta}+b^2\sin^2{\beta}}{a^4\cos^2{\alpha}\cos^2{\beta}+b^4\sin^2{\alpha}\sin^2{\beta}+a^2b^2\left(\sin^2{\alpha}\cos^2{\beta}+\cos^2{\alpha}\sin^2{\beta}\right)}\\&= \dfrac{\left(a^2+b^2\right)\sin^2{\left(\alpha-\beta\right) } }{a^2b^2\sin^2{\left(\alpha-\beta\right)}} \\&= \dfrac{1}{a^2}+\dfrac{1}{b^2}. \end{split}\]又易证 $\dfrac{1}{d^2}=\dfrac{1}{r_1^2}+\dfrac{1}{r_2^2}$，故$$\dfrac{1}{d^2}=\dfrac{1}{r_1^2}+\dfrac{1}{r_2^2}=\dfrac{1}{a^2}+\dfrac{1}{b^2}.$$反之，若 $\dfrac{1}{d^2}=\dfrac{1}{r_1^2}+\dfrac{1}{r_2^2}=\dfrac{1}{a^2}+\dfrac{1}{b^2}$，则类似可证 $OA\perp OB$．
证毕．
若 $OA\perp OB$，设点 $A$ 坐标为 $\left(r_1\cos \theta,r_1 \sin \theta \right)$，则点 $B$ 坐标为 $\left(r_2\cos \left(\theta \pm \dfrac{\pi}{2}\right),r_2 \sin \left(\theta \pm \dfrac{\pi}{2}\right) \right)$．
将 $A,B$ 坐标代入椭圆方程，可得$$\begin{cases}\dfrac{1}{r_1^2}&=\dfrac{\cos ^2 \theta}{a^2}+\dfrac{\sin ^2 \theta}{b^2},\\ \dfrac{1}{r_2^2}&=\dfrac{\sin ^2 \theta}{a^2}+\dfrac{\cos ^2 \theta}{b^2}. \end{cases}$$以上两式相加，即得 $\dfrac{1}{r_1^2}+\dfrac{1}{r_2^2}=\dfrac{1}{a^2}+\dfrac{1}{b^2}$．又易证 $\dfrac{1}{d^2}=\dfrac{1}{r_1^2}+\dfrac{1}{r_2^2}$，故$$\dfrac{1}{d^2}=\dfrac{1}{r_1^2}+\dfrac{1}{r_2^2}=\dfrac{1}{a^2}+\dfrac{1}{b^2}.$$反之，若 $\dfrac{1}{d^2}=\dfrac{1}{r_1^2}+\dfrac{1}{r_2^2}=\dfrac{1}{a^2}+\dfrac{1}{b^2}$，则类似可证 $OA\perp OB$．
证毕．
{\bf{方法二}}
设点 $A$ 坐标为 $\left(a\cos \alpha,b \sin \alpha \right)$，点 $B$ 坐标为 $\left(a\cos \beta,b \sin \beta \right)$．因为 $OA\perp OB$，所以$$a^2\cos{\alpha}\cos{\beta}+b^2\sin{\alpha}\sin{\beta}=0.$$因为\[\begin{split}a^2\cos^2{\alpha}&=a^2\cos^2{\alpha}\sin^2{\beta}+a^2\cos^2{\alpha}\cos^2{\beta},\\b^2\sin^2{\alpha}&=b^2\sin^2{\alpha}\sin^2{\beta}+b^2\sin^2{\alpha}\cos^2{\beta},\\a^2\cos^2{\beta}&=a^2\cos^2{\beta}\sin^2{\alpha}+a^2\cos^2{\beta}\cos^2{\alpha},\\b^2\sin^2{\beta}&=b^2\sin^2{\beta}\sin^2{\alpha}+b^2\sin^2{\beta}\cos^2{\alpha},\end{split}\]所以\[\begin{split}&a^2\cos^2{\alpha}+b^2\sin^2{\alpha}+a^2\cos^2{\beta}+b^2\sin^2{\beta}\\=& \left(a^2+b^2\right)\left(\sin^2{\alpha}\cos^2{\beta}+\cos^2{\alpha}\sin^2{\beta}\right)+2a^2\cos^2{\alpha}\cos^2{\beta}+2b^2\sin^2{\alpha}\sin^2{\beta}\\=&\left(a^2+b^2\right)\left(\sin^2{\alpha}\cos^2{\beta}+\cos^2{\alpha}\sin^2{\beta}\right)-2b^2\sin{\alpha}\sin{\beta}cos{\alpha}\cos{\beta}-2a^2\sin{\alpha}\sin{\beta}cos{\alpha}\cos{\beta}\\=&\left(a^2+b^2\right)\left(\sin{\alpha}\cos{\beta}-\cos{\alpha}\sin{\beta}\right)^2\\=&\left(a^2+b^2\right)\sin^2{\left(\alpha-\beta\right) }. \end{split}\]因为\[a^4\cos^2{\alpha}\cos^2{\beta}+b^4\sin^2{\alpha}\sin^2{\beta}=-2a^2b^2\sin{\alpha}\cos{\beta}\cos{\alpha}\sin{\beta},\]所以\[\begin{split}&a^4\cos^2{\alpha}\cos^2{\beta}+b^4\sin^2{\alpha}\sin^2{\beta}+a^2b^2\left(\sin^2{\alpha}\cos^2{\beta}+\cos^2{\alpha}\sin^2{\beta}\right)\\=&a^2b^2\sin^2{\left(\alpha-\beta\right)}. \end{split}\]故\[\begin{split}\dfrac{1}{r_1^2}+\dfrac{1}{r_2^2}&=\dfrac{1}{a^2\cos^2{\alpha}+b^2\sin^2{\alpha}}+\dfrac{1}{a^2\cos^2{\beta}+b^2\sin^2{\beta}}\\&= \dfrac{a^2\cos^2{\alpha}+b^2\sin^2{\alpha}+a^2\cos^2{\beta}+b^2\sin^2{\beta}}{a^4\cos^2{\alpha}\cos^2{\beta}+b^4\sin^2{\alpha}\sin^2{\beta}+a^2b^2\left(\sin^2{\alpha}\cos^2{\beta}+\cos^2{\alpha}\sin^2{\beta}\right)}\\&= \dfrac{\left(a^2+b^2\right)\sin^2{\left(\alpha-\beta\right) } }{a^2b^2\sin^2{\left(\alpha-\beta\right)}} \\&= \dfrac{1}{a^2}+\dfrac{1}{b^2}. \end{split}\]又易证 $\dfrac{1}{d^2}=\dfrac{1}{r_1^2}+\dfrac{1}{r_2^2}$，故$$\dfrac{1}{d^2}=\dfrac{1}{r_1^2}+\dfrac{1}{r_2^2}=\dfrac{1}{a^2}+\dfrac{1}{b^2}.$$反之，若 $\dfrac{1}{d^2}=\dfrac{1}{r_1^2}+\dfrac{1}{r_2^2}=\dfrac{1}{a^2}+\dfrac{1}{b^2}$，则类似可证 $OA\perp OB$．
证毕．
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