若点P是椭圆C:x2a2+y2b2=1(a>b>0)上一点,F1,F2是左右焦点,求三角形PF1...
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发布时间:2024-10-21 05:02
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时间:2024-11-07 19:26
∵a>b>0,
∴c2=a2-b2,
∴c=a2?b2;
设△PF1F2内切圆半径为r,
则S△PF1F2=r2(PF1+PF2+F1F2)
=r2(2a+2c)
=r(a+c)
=r(a+a2?b2),
显然,当三角形PF1F2的面积最大时,半径最大,当点P为上端点或下端点时,面积最大,为bc=ba2?b2,
∴rmax=bca+c
=ba2?b2a+a2?b2
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简单计算一下,答案如图所示
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时间:2024-11-07 19:24
简单计算一下,答案如图所示