Question

抛物线的焦点F在x轴上，A(m,－3)在抛物线上，且|AF|=5，求抛物线的标准方程.

Answer 1

解:设抛物线方程为y²=2px或y²=－2px(p＞0).

∵A点在抛物线上,

∴(－3)²=2pm或(－3)²=－2pm.

∴m=±.                                                                                      ①

又|AF|=+|m|=5,                                                                               ②

把①代入②可得+=5,

即p²－10p+9=0.∴p=1或p=9.

∴所求抛物线方程为y²=±2x或y²=±18x.