設(shè)P（x1,y1）Q（x2,y2）
根據(jù)題意y1/x1*y2/x2=-1
即x1x2+y1y2=0
設(shè)PQ方程：y=kx+m代入橢圓b²x²+a²y²=a²b²
整理：(a²k²+b²)x²+2kma²x+a²m²-a²b²=0
韋達(dá)定理：x1+x2=-2kma²/(a²k²+b²),x1*x2=(a²m²-a²b²)/(a²k²+b²)
y1y2=(kx1+m)(kx2+m)=k²x1x2+km(x1+x2)+m²
x1x2+k²x1x2+km(x1+x2)+m²=0
(a²m²-a²b²)/(a²k²+b²)+k²(a²m²-a²b²)/(a²k²+b²)-2k²m²a²/(a²k²+b²)+m²=0
化簡(jiǎn)：(a²+b²)m²=a²b²(1+k²)
m²/(1+k²)=a²b²/(a²+b²)
｜m｜/√(1+k²)=ab/√(a²+b²)
點(diǎn)O到直線PQ的距離d=｜m｜/√(1+k²)=ab/√(a²+b²)為定值
1/OP²+1/OQ²=(OP²+OQ²)/(OP²*OQ²)=PQ²/(PQ*d)²=1/d²=1/[a²b²/(a²+b²)]
=(a²+b²)/(a²b²)=1/a²+1/b²