A,B是椭圆x^2/a^2+y^2/b^2=1上两点,且OA垂直OB,求证1/OA^2+1/OB^2为定值
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A,B是椭圆x^2/a^2+y^2/b^2=1上两点,且OA垂直OB,求证1/OA^2+1/OB^2为定值
A,B是椭圆x^2/a^2+y^2/b^2=1上两点,且OA垂直OB,求证1/OA^2+1/OB^2为定值
A,B是椭圆x^2/a^2+y^2/b^2=1上两点,且OA垂直OB,求证1/OA^2+1/OB^2为定值
将椭圆方程改写为:x=acosθ,y=bsinθ,其中θ为OP(x,y)与Ox轴的夹角
设A(x1,y1)对应的是θ1,B(x2,y2)对应的是θ2
根据题意,OA⊥OB,则|θ2-θ1|=π/2
不失一般性,可另θ2=θ1+π/2
则cosθ2=-sinθ1,sinθ2=cosθ1
x1 = acosθ1,y1 = bsinθ1;
x2 = acosθ2 = -asinθ1,y2 = bsinθ2 = bcosθ1
|OA|^2 = x1^2 + y1^2 = a^2cos^2θ1 + b^2sin^2θ1
|OB|^2 = x2^2 + y2^2 = a^2sin^2θ1 + b^2cos^2θ1
|OA|^2+|OB|^2 = (a^2+b^2)*(cos^2θ1+sin^2θ1) = a^2+b^2
|OA|^2*|OB|^2 = (a^2cos^2θ1 + b^2sin^2θ1)*(a^2sin^2θ1 + b^2cos^2θ1)
= (a^4+b^4)*sin^2θ1cos^2θ1 + a^2b^2*(cos^4θ1+sin^4θ1)
= (a^4+b^4-2a^2b^2)*sin^2θ1cos^2θ1 + a^2b^2*(cos^4θ1+sin^4θ1+2sin^2θ1cos^2θ1)
= (a^2-b^2)^2*sin^2θ1cos^2θ1 + a^2b^2*(cos^2θ1+sin^2θ1)^2
= (a^2-b^2)^2*sin^2θ1cos^2θ1 + a^2b^2
= (ab)^2 + (c*sinθ1cosθ1)^2
1/|OA|^2 + 1/|OB|^2 = (|OA|^2 + |OB|^2)/(|OA|^2*|OB|^2)
= (a^2+b^2)/[(ab)^2+(c*sinθ1cosθ1)^2]
似乎不为常数嘛
可以设A(acosθ,bsinθ)B(acosα,bsinα)其中α=θ+π/2 则有OA^2+OB^2=a^2cosθ^2+b^2sinθ^2+a^2cosα^2+b^2sinα^2 =a^2+b^2 OA^2×OB^2=(a^2cosθ^2+b^2sinθ^2)×(a^2cosα^2+b^2sinα^2) =[a^2(1-sinθ^2)+b^2sinθ^2]×[a^2sinθ^2+b^2(1-sinθ^2)] =(a^2-sinθ^2c^2)(b^2+c^2sinθ^2) =a^2b^2+a^2c^2sinθ^2-b^2c^2sinθ^2-c^4sinθ^2 =a^2b^2 可证得:1/OA^2+1/OB^2 =(OA^2+OB^2)/(OA^2×OB^2) =(a^2+b^2)/(a^2b^2) =1/a^2+1/b^2 看不懂欢迎追问 求采纳