一道关于椭圆定值的问题!（难）椭圆中心O,长轴,短轴分别为2a,2b,A.B分别为椭圆的两点,OA垂直OB,求证：1/O
1个回答
该椭圆方程为
x^2/a^2+y^2/b^2=1
可设点A坐标为(acosα,bsinα),B坐标为(acosβ,bsinβ)
OA垂直OB,所以acosα*acosβ+bsinα*bsinβ=0
两遍同除cosα*cosβ,得
a^2+b^2tanα*tanβ=0
tanβ=-(a^2/(b^2*tanα))
1/OA^2+1/OB^2
=1/(a^2(cosα)^2+b^2(sinα)^2) + 1/(a^2(cosβ)^2+b^2(sinβ)^2)
=((cosα)^2+(sinα)^2)/(a^2(cosα)^2+b^2(sinα)^2) + ((cosβ)^2+(sinβ)^2)/(a^2(cosβ)^2+b^2(sinβ)^2)
=(1+(tanα)^2)/(a^2+b^2(tanα)^2) + (1+(tanβ)^2)/(a^2+b^2(tanβ)^2)
=(1+(tanα)^2)/(a^2+b^2(tanα)^2) + (1+((a^2/(b^2tanα)))^2)/(a^2+b^2((a^2/(b^2tanα)))^2)
=(a^2*b^2+a^2*b^2*(tanα)^2+b^4*(tanα)^2+a^4)/(a^2*b^4*(tanα)^2+b^2*a^4)
=(a^2+b^2)/(a^2*b^2)
所以1/OA^2+1/OB^2为定值(a^2+b^2)/(a^2*b^2)
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