Question

3£®ÒÑ֪˫ÇúÏßC£º$\frac{x^2}{a^2}$-$\frac{y^2}{b^2}$=1£¨a£¾0£¬b£¾0£©µÄÀëÐÄÂÊΪ$\sqrt{3}$£¬ÓÒ×¼Ïß·½³ÌΪx=$\frac{{\sqrt{3}}}{3}$
£¨¢ñ£©ÇóË«ÇúÏßCµÄ·½³Ì£»
£¨¢ò£©ÉèÖ±ÏßlÊÇÔ²O£ºx²+y²=r²É϶¯µãP£¨x₀£¬y₀£©£¨x₀y₀¡Ù0£©´¦µÄÇÐÏߣ¬lÓëË«ÇúÏßC½»ÓÚ²»Í¬µÄÁ½µãA£¬B£¬ÊÇ·ñ´æÔÚʵÊýrʹµÃ¡ÏAOBʼÖÕΪ90¡ã£®Èô´æÔÚ£¬Çó³örµÄÖµ£»Èô²»´æÔÚ£¬ËµÃ÷ÀíÓÉ£®

Answer 1

·ÖÎö £¨¢ñ£©ÀûÓÃÀëÐÄÂÊΪ$\sqrt{3}$£¬ÓÒ×¼Ïß·½³ÌΪx=$\frac{{\sqrt{3}}}{3}$£¬Áгö·½³Ì×飬Çó³öa£¬c£¬b£¬¼´¿ÉÇó½âË«ÇúÏߵķ½³Ì£®
£¨¢ò£©µãP£¨x₀£¬y₀£©£¨x₀y₀¡Ù0£©ÔÚÔ²x²+y²=r²ÉÏ£¬µÃµ½ÇÐÏß·½³Ì£¬ÓëË«ÇúÏßÁªÁ¢£¬ÀûÓÃΤ´ï¶¨Àí½áºÏÏòÁ¿µÄÊýÁ¿»ý£¬Çó½â¼´¿É£®

½â´ð ½â£º£¨¢ñ£©ÓÉÌâÒ⣬µÃ$\left\{{\begin{array}{l}{\frac{a^2}{c}=\frac{{\sqrt{3}}}{3}}\\{\frac{c}{a}=\sqrt{3}}\end{array}}\right.$£¬½âµÃ$a=1£¬c=\sqrt{3}$£¬
¡àb²=c²-a²=2£¬¡àËùÇóË«ÇúÏßCµÄ·½³ÌΪ${x^2}-\frac{y^2}{2}=1$£®¡­..£¨4·Ö£©
£¨¢ò£©µãP£¨x₀£¬y₀£©£¨x₀y₀¡Ù0£©ÔÚÔ²x²+y²=r²ÉÏ£¬
Ô²ÔÚµãP£¨x₀£¬y₀£©´¦µÄÇÐÏß·½³ÌΪ$y-{y_0}=-\frac{x_0}{y_0}£¨x-{x_0}£©$£¬
»¯¼òµÃ$x{x_0}+y{y_0}={r^2}$£®¡­..£¨5·Ö£©
ÓÉ$\left\{{\begin{array}{l}{{x^2}-\frac{y^2}{2}=1}\\{x{x_0}+y{y_0}={r^2}}\end{array}}\right.$ÏûÈ¥yµÃ$£¨2y_0^2-x_0^2£©{x^2}+2{r^2}{x_0}x-{r^4}-2y_0^2=0$¢Ù$£¨2y_0^2-x_0^2£©{y^2}-4{r^2}{y_0}y+2{r^4}-2x_0^2=0$¢Ú¡­..£¨8·Ö£©
Èô´æÔÚʵÊýr Ê¹µÃ¡ÏAOBʼÖÕΪ90⁰ÔòÓÐ$\overrightarrow{OA}•\overrightarrow{OB}={x_1}{x_2}+{y_1}{y_2}=0$£¬
¶ø${x_1}{x_2}=\frac{{-{r^4}-2y_0^2}}{2y_0^2-x_0^2}$£¬${y_1}{y_2}=\frac{{2{r^4}-2x_0^2}}{2y_0^2-x_0^2}$ÓÖ$x_0^2+y_0^2={r^2}$£¬
x₁x₂+y₁y₂=$\frac{{-{r^4}-2y_0^2}}{2y_0^2-x_0^2}+$$\frac{{2{r^4}-2x_0^2}}{2y_0^2-x_0^2}$=$\frac{{{r^4}-2{r^2}}}{2y_0^2-x_0^2}$=0£¬
$r=\sqrt{2}$¡­..£¨10·Ö£©
¶ø$r=\sqrt{2}$ʱ¢Ù»¯Îª$£¨3x_0^2-4£©{x^2}-4{x_0}x+8-2x_0^2=0$£¬x₀y₀¡Ù0£¬
$0£¼x_0^2£¼2$£¬
$¡÷=16x_0^2-4£¨3x_0^2-4£©£¨8-2x_0^2£©£¾0$£¬
×ÛÉÏËùÊö´æÔÚ$r=\sqrt{2}$ʹµÃ¡ÏAOBʼÖÕΪ90¡ã¡­..£¨12·Ö£©

µãÆÀ ±¾Ì⿼²éË«ÇúÏß·½³ÌµÄÇ󷨣¬Ö±ÏßÓëË«ÇúÏßµÄλÖùØϵµÄ×ÛºÏÓ¦Ó㮿¼²é´æÔÚÐÔÎÊÌâµÄÇó½â·½·¨£¬¿¼²é·ÖÎöÎÊÌâ½â¾öÎÊÌâµÄÄÜÁ¦£¬×ª»¯Ë¼ÏëµÄÓ¦Óã®