解:(1) 由已知 ,a2+b2=5.又a2= b2+c2,解得a2=4,b2=1, 所以椭圆C的方程为![](http://img.shitiku.com.cn/uploads/allimg/20191023/20191023111119-13546.png) (2)根据题意,过点D(0,4)满足题意的直线斜率存在, 设l:y= kx +4. 联立 消去y得(1+4k2)x2+32kx+60=0. 由题知Δ=(32k)2-240(1+4k2)=64k2-240>0,解得![](http://img.shitiku.com.cn/uploads/allimg/20191023/20191023111120-16031.png) 设E,F两点的坐标分别为(x1,y1),(x2,y2), 则![](http://img.shitiku.com.cn/uploads/allimg/20191023/20191023111120-49941.png) ![](http://img.shitiku.com.cn/uploads/allimg/20191023/20191023111120-99938.png) 因为OE⊥OF,所以 ,即x1x2+y1y2=0, 所以(1+k2)x1x2+4k(x1+x2)+16=0, 所以 ,解得![](http://img.shitiku.com.cn/uploads/allimg/20191023/20191023111121-88960.png) 所以直线l的斜率为![](http://img.shitiku.com.cn/uploads/allimg/20191023/20191023111121-10113.png) |