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