解:(1)设椭圆C的方程为![](http://img.shitiku.com.cn/uploads/allimg/20191023/20191023122400-35998.png) ∵椭圆的一个顶点恰好是抛物线y2= 的焦点, ∴a=![](http://img.shitiku.com.cn/uploads/allimg/20191023/20191023122401-88922.png) ∵离心率等于 , ∴ , ∴c=1∴b=1 ∴椭圆C的方程为 ; (2)设A(x1,y1),B(x2,y2),直线AB的方程为y=x+t, 代入椭圆方程,消元可得3x2+4tx+2t2﹣2=0 由△>0,解得﹣ <t<![](http://img.shitiku.com.cn/uploads/allimg/20191023/20191023122402-76538.png) 由韦达定理得x1+x2=﹣ t,x1x2= . ∵PQ过椭圆焦点且PQ⊥x轴, ∴|PQ|=![](http://img.shitiku.com.cn/uploads/allimg/20191023/20191023122403-29323.png) ∴四边形APBQ的面积S= × ×|x1﹣x2|= ×![](http://img.shitiku.com.cn/uploads/allimg/20191023/20191023122404-69613.png) ∴t=0时,Smax= ; (3)当∠APQ=∠BPQ,则PA、PB的斜率之和为0, 设直线PA的斜率为k,则PB的斜率为﹣k, PA的直线方程为y﹣ =k(x﹣1), 与椭圆方程联立,消元可得(1+2k2)x2+(2 k﹣4k2)x+k2﹣2 k﹣1=0 ∴x1+1=﹣![](http://img.shitiku.com.cn/uploads/allimg/20191023/20191023122405-14373.png) 同理x2+1=﹣![](http://img.shitiku.com.cn/uploads/allimg/20191023/20191023122405-78517.png) ∴x1+x2= ,x1﹣x2=﹣![](http://img.shitiku.com.cn/uploads/allimg/20191023/20191023122406-23704.png) ∴y1﹣y2=k(x1+x2)﹣2k= ,x1﹣x2=﹣![](http://img.shitiku.com.cn/uploads/allimg/20191023/20191023122406-65834.png) ∴![](http://img.shitiku.com.cn/uploads/allimg/20191023/20191023122406-86932.png) ∴直线AB的斜率为定值 . |