如题所述
设F₁ï¼F₂æ¯æ¤åX²/25+Y²/16=1ç两个ç¦ç¹ï¼På¨æ¤åä¸ï¼ä¸â F₁PF₂=60°ï¼æ±â³F₁PF₂çé¢ç§¯ï¼
解ï¼a=5ï¼b=4ï¼c=3ï¼F₁(-3ï¼0)ï¼F₂(3ï¼0)ï¼
å¨â³F₁PF₂ä¸ï¼F₁F₂ï¼6ï¼â F₁PF₂=60°ï¼è®¾PF₁=Ï₁ï¼PF₂=Ï₂ï¼Ï₁+Ï₂=10ï¼åç±ä½å¼¦å®çå¾ï¼
Ï₁²+Ï₂²-2Ï₁Ï₂cos60°=Ï₁²+Ï₂²-Ï₁Ï₂=(Ï₁+Ï₂)²-3Ï₁Ï₂=100-3Ï₁Ï₂=36ï¼æ Ï₁Ï₂=64/3ï¼
â´â³F₁PF₂çé¢ç§¯=(1/2)Ï₁Ï₂sin60°=(1/2)Ã(64/3)Ã(â3/2)=(16/3)â3.
解ï¼a=5ï¼b=4ï¼c=3ï¼F₁(-3ï¼0)ï¼F₂(3ï¼0)ï¼
å¨â³F₁PF₂ä¸ï¼F₁F₂ï¼6ï¼â F₁PF₂=60°ï¼è®¾PF₁=Ï₁ï¼PF₂=Ï₂ï¼Ï₁+Ï₂=10ï¼åç±ä½å¼¦å®çå¾ï¼
Ï₁²+Ï₂²-2Ï₁Ï₂cos60°=Ï₁²+Ï₂²-Ï₁Ï₂=(Ï₁+Ï₂)²-3Ï₁Ï₂=100-3Ï₁Ï₂=36ï¼æ Ï₁Ï₂=64/3ï¼
â´â³F₁PF₂çé¢ç§¯=(1/2)Ï₁Ï₂sin60°=(1/2)Ã(64/3)Ã(â3/2)=(16/3)â3.
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第1个回答 2013-01-27
解:X^2/25+Y^2/16=1中
a=5,b=4
所以c=3
由PF1+PF2=2a=10
得(PF1)^2+2(PF1)(PF2)+(PF2)^2=100①
又(PF1)^2-2(PF1)(PF2)cos60°+(PF2)^2=(F1F2)^2=6^2
得(PF1)^2-(PF1)(PF2)+(PF2)^2=36②
①-②得3(PF1)(PF2)=64
得(PF1)(PF2)=64/ 3
S△PF1F2
=(1/2)|PF1||PF2|sin60°
=(1/2)[64/2*(√3/2)
=8√3)
a=5,b=4
所以c=3
由PF1+PF2=2a=10
得(PF1)^2+2(PF1)(PF2)+(PF2)^2=100①
又(PF1)^2-2(PF1)(PF2)cos60°+(PF2)^2=(F1F2)^2=6^2
得(PF1)^2-(PF1)(PF2)+(PF2)^2=36②
①-②得3(PF1)(PF2)=64
得(PF1)(PF2)=64/ 3
S△PF1F2
=(1/2)|PF1||PF2|sin60°
=(1/2)[64/2*(√3/2)
=8√3)
第2个回答 2013-01-27
s△=b^2tan30°=16√3/3