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