Question

已知椭圆的中心为O,长轴.短轴的长分别为2a,2b(a>b>0),A,B分别为椭圆上的两

Answer 1

设OA的所在直线方程为y=kx,则OB所在直线方程为y=-x/k; 它们与椭圆的交点A、B坐标（xa,ya)、(xb,yb)满足 xa^2=1/[1/a^2+k^2/b^2] ya^2=k^2/[1/a^2+k^2/b^2] xb^2=1/[1/a^2+1/(k^2b^2)] yb^2=1/[k^2/a^2+1/b^2] OA^2=xa^2+ya^2=(1+k^2)/[1/a^2+k^2/b^2] OB^2=xb^2+yb^2=(1+1/k^2)/[1/a^2+1/(k^2b^2)] 1/OA^2+1/OB^2=[1/a^2+k^2/b^2]/(1+k^2)+[1/a^2+1/(k^2b^2)]*k^2/(1+k^2) =1/a^2+1/b^2为定值。 ======================================================= 以中心为极点，x轴为极轴建立极坐标系 方程为ρ^2(cosθ)^2/a^2+ρ^2(sinθ)^2/b^2=1 1/ρ^2=(cosθ)^2/a^2+(sinθ)^2/b^2 设A(ρ1,θ)，由OA⊥OB得B(ρ2,θ+π/2) 1/OA^2+1/OB^2＝1/ρ1^2+1/ρ2^2 =(cosθ)^2/a^2+(sinθ)^2/b^2+(cos(θ+π/2))^2/a^2+(sin(θ+π/2))^2/b^2 =(cosθ)^2/a^2+(sinθ)^2/b^2+(sinθ)^2/a^2+(cosθ)^2/b^2