每日一题[821]百花齐放

已知$A,B$是椭圆$E:\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}=1$($a>b>0$)上的两点，$O$为坐标原点，且$OA\perp OB$，求证：$O$到直线$AB$的距离为定值．

分析与解　法一　直线消参

当直线$AB$的斜率存在时，设其方程为$y=kx+m$，$A(x_1,y_1)$，$B(x_2,y_2)$．根据题意，有\[x_1x_2+y_1y_2=0,\]即\[\left(k^2+1\right)x_1x_2+km(x_1+x_2)+m^2=0.\]联立直线与椭圆方程，可得\[\left(\dfrac{1}{a^2}+\dfrac{k^2}{b^2}\right)x^2+\dfrac{2km}{b^2}x+\dfrac{m^2}{b^2}-1=0,\]这样就有\[\left(k^2+1\right)\left(\dfrac{m^2}{b^2}-1\right)+km\left(-\dfrac{2km}{b^2}\right)+m^2\left(\dfrac 1{a^2}+\dfrac{k^2}{b^2}\right)=0,\]也即\[1+k^2=\left(\dfrac{1}{a^2}+\dfrac{1}{b^2}\right)m^2.\]因此$O$到直线$AB$的距离\[d=\dfrac{|m|}{\sqrt{1+k^2}}=\dfrac{ab}{\sqrt{a^2+b^2}}\]为定值．

经验证，当直线$AB$的斜率不存在时，也有$d=\dfrac{ab}{\sqrt{a^2+b^2}}$，因此原命题得证．

法二　化齐次联立

设直线$AB$的方程为$mx+ny=1$，与椭圆方程化齐次联立，可得\[\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}-(mx+ny)^2=0,\]从而由$OA\perp OB$可得\[\dfrac{1}{a^2}+\dfrac{1}{b^2}-m^2-n^2=0,\]因此$O$到直线$AB$的距离\[d=\dfrac{1}{\sqrt{m^2+n^2}}=\dfrac{ab}{\sqrt{a^2+b^2}}\]为定值．

法三　相关直线

当直线$OA$和直线$OB$的斜率均存在时，设$OA:y=kx$，$OB:y=-\dfrac 1kx$．联立直线$OA$与椭圆的方程，可得\[\left(\dfrac{1}{a^2}+\dfrac{k^2}{b^2}\right)x^2=1.\]于是$O$到直线$AB$的距离$d$满足
\[\begin{split}\dfrac{1}{d^2}&=\dfrac{1}{|OA|^2}+\dfrac{1}{|OB|^2}\\
&=\dfrac{\dfrac{1}{a^2}+\dfrac{k^2}{b^2}}{1+k^2}+\dfrac{\dfrac{1}{a^2}+\dfrac{1}{b^2k^2}}{1+\dfrac{1}{k^2}}\\
&=\dfrac{1}{a^2}+\dfrac{1}{b^2},\end{split}\]因此$d$为定值$\dfrac{ab}{\sqrt{a^2+b^2}}$．

经验证，当直线$OA$或$OB$的斜率不存在时，也有$d=\dfrac{ab}{\sqrt{a^2+b^2}}$，因此原命题得证．

法四　椭圆消参

设$A(a\cos\alpha,b\sin\alpha)$，$B(a\cos\beta,b\sin\beta)$，则有由$OA\perp OB$有\[a^2\cos\alpha\cos\beta+b^2\sin\alpha\sin\beta=0,\]当$\cos\alpha\cos\beta\ne 0$时，有\[\tan^2\alpha\cdot \tan^2\beta=\dfrac{a^4}{b^4}.\]此时$O$到直线$AB$的距离$d$满足\[\begin{split}\dfrac{1}{d^2}&=\dfrac{1}{|OA|^2}+\dfrac{1}{|OB|^2}\\&=\dfrac{1}{a^2\cos^2\alpha+b^2\sin^2\alpha}+\dfrac{1}{a^2\cos^2\beta+b^2\sin^2\beta}\\
&=\dfrac{\sin^2\alpha+\cos^2\alpha}{a^2\cos^2\alpha+b^2\sin^2\alpha}+\dfrac{\sin^2\beta+\cos^2\beta}{a^2\cos^2\beta+b^2\sin^2\beta}\\&=\dfrac{\tan^2\alpha+1}{a^2+b^2\tan^2\alpha}+\dfrac{\tan^2\beta+1}{a^2+b^2\tan^2\beta}\\
&=\dfrac{\tan^2\alpha+1}{a^2+b^2\tan^2\alpha}+\dfrac{\left(\tan^2\beta+1\right)\cdot\tan^2\alpha}{\left(a^2+b^2\tan^2\beta\right)\cdot \tan^2\alpha}\\&=\dfrac{\tan^2\alpha+1}{a^2+b^2\tan^2\alpha}+\dfrac{\dfrac{a^4}{b^4}+\tan^2\alpha}{a^2\tan^2\alpha+b^2\cdot \dfrac{a^4}{b^4}}\\&=\dfrac{\tan^2\alpha+1}{a^2+b^2\tan^2\alpha}+\dfrac{\dfrac{a^2}{b^2}+\dfrac{b^2}{a^2}\tan^2\alpha}{b^2\tan^2\alpha+a^2}\\&=\dfrac{1}{a^2}+\dfrac{1}{b^2},\end{split}\]因此$d$为定值$\dfrac{ab}{\sqrt{a^2+b^2}}$．

经验证，当$\cos\alpha\cos\beta=0$时，$d=\dfrac{ab}{\sqrt{a^2+b^2}}$，因此原命题得证．

法五　极坐标

设$A\left(\theta:r_1\right)$，$B\left(\theta+\dfrac{\pi}2,r_2\right)$，则有\[\begin{aligned}\dfrac{r_1^2\cos^2\theta}{a^2}+\dfrac{r_1^2\sin^2\theta}{b^2}&=1,\\\dfrac{r_2^2\cos^2\left(\theta+\dfrac{\pi}2\right)}{a^2}+\dfrac{r_2^2\sin^2\left(\theta+\dfrac{\pi}2\right)}{b^2}&=1,
\end{aligned}\]于是$O$到直线$AB$的距离$d$满足\[\dfrac{1}{d^2}=\dfrac{1}{r_1^2}+\dfrac{1}{r_2^2}=\dfrac{1}{a^2}+\dfrac{1}{b^2}\]为定值，因此原命题得证．