1. 证明：∵直线l:mx+ny=1与椭圆C:x^2/a^2+y^2/b^2=1(a＞b＞0)交于不同两点
联立方程得：mx+ny=1 ①
:x^2/a^2+y^2/b^2=1 ②
化简整理得：（b^2+a^2m^2/n^2）x^2-2a^2m/n^2x+a^2/n^2-a^2b^2=0
由根的判别式△>0得4a^4m^2/n^2-4(b^2+a^2m^2/n^2)(a^2/n^2-a^2b^2)>0
化简得(a^2b^4n^2-a^2b^2+a^4b^2m^2)/n^2>0
因为n^2>0
所以a^2b^4n^2-a^2b^2+a^4b^2m^2>0
即a^2b^4n^2+a^4b^2n^2>a^2b^2
也即a^2m^2+b^2n^2>1
2. 证明：设P(x1,y1),R(x2,y2)
因为OP⊥OR
所以向量OP.向量OR=0
即x1x2+y1y2=0
由韦达定理得 x1+x2=2a^2m/(b^2n^2+a^2m^2)
X1.x2=(a^2-a^2b^2n^2)/(b^2n^2+a^2m^2)
带入mx+ny=1得 y1.y2=(b^2n^2-a^2b^2n^2m^2)/(b^2n^4+a^2m^2n^2)
∴(a^2-a^2b^2n^2)/(b^2n^2+a^2m^2)+ (b^2n^2-a^2b^2n^2m^2)/(b^2n^4+a^2m^2n^2)=0
即(a^2+b^2-a^2b^2n^2-a^2b^2m^2)/(b^2n^2+a^2m^2)=0
也即a^2+b^2=a^2b^2(m^2+n^2)