æ±æ¤å 3x'²-2(â3)x'y'+5y'²-1=0ç离å¿çã
解ï¼ä»¤x'=xcosα-ysinÎ±ï¼ y'=xsinα+ycosαï¼ä»£å
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3(xcosα-ysinα)²-(2â3)(xcosα-ysinα)(xsinα+ycosα)+5(xsinα+ycosα)²-1=0ï¼
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[3cos²Î±+5sin²Î±-2(â3)sinαcosα]x²+[3sin²Î±+5cos²Î±-(2â3)sinαcosα)]y²-1
+[4sinαcosα+(2â3)sin²Î±-(2â3)cos²Î±]xy=0............â
为äºæ¶å»xy项ï¼ä»¤4cosαsinα+(2â3)sin²Î±-(2â3)cos²Î±=2sin2α+(2â3)cos2α=0
äºæ¯å¾tan2α=-â3ï¼æ
å¾2α=Ï-Ï/3=2Ï/3ï¼â´Î±=Ï/3ï¼
å³å°åæ è½´æ顺æ¶éæ¹åæ转Ï/3å°±è½æ¶å»äº¤å项xy,å°Î±=Ï/3代å
¥â å¼ï¼äºæ¯åæ¹ç¨å
æäºï¼(3/4+15/4-3/2)x²+(9/4+5/4-3/2)y²-1=0
å³æ 3x²+2y²=1ï¼å³x²/(1/3)+y²/(1/2)=1ï¼è¿å°±æ¯è¯¥æ¤åçæ åæ¹ç¨ï¼ç±æ¤å¯ç¥ï¼
a²=1/2ï¼b²=1/3ï¼c²=1/2-1/3=1/6ï¼æ
离å¿çe=c/a=â(1/3).
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