![](http://img.shitiku.com.cn/uploads/allimg/20191019/20191019090825-93912.jpg) 所以抛物线的解析式是y=x2+2x-3
![](http://img.shitiku.com.cn/uploads/allimg/20191019/20191019090826-86589.jpg) (2)解法一:假设抛物线上存在点G,设G(m,n),显然,当n=-3时,△AGH不存在。
![](http://img.shitiku.com.cn/uploads/allimg/20191019/20191019090826-55699.jpg) S△AGH= S△GHC,∴m+n+1=0,
∵点G在y轴的左侧,∴G(-1,-4).
![](http://img.shitiku.com.cn/uploads/allimg/20191019/20191019090826-41367.jpg) 解法二:①如图①,当GH//AC时,点A,点C到GH的距离相等,所以S△AGH= S△GHC,可得AC的解析式为y=3x-3,∵GH//AC,得GH的解析式为y=3x-1. ∴G(-1,-4) ②如图②,当GH与AC不平行时,因为点A,C到直线GH的距离相等,所以直线GH
![](http://img.shitiku.com.cn/uploads/allimg/20191019/20191019090827-22844.jpg) (3) 如图③,E(-2,0), ∴D点的横坐标是-2,点D在抛物线上,∴D(-2,-3)
![](http://img.shitiku.com.cn/uploads/allimg/20191019/20191019090827-67328.jpg) ∠BPE+∠EPF+∠FPD=∠DFP+∠PDF+∠FPD=180°, ∠EPF=∠PDF,∴∠BPE=∠DFP,可证△PBE∽△FDP,
![](http://img.shitiku.com.cn/uploads/allimg/20191019/20191019090827-85669.jpg) |