Question

抛物线的焦点F在X轴上,直线Y=﹣3与抛物线相交于点A,|AF|=5,求抛物线的标准方程

Answer 1

设 抛物线线的方程为 y^2=4mx

则 焦点F(m,0),准线：x=-m

(-3)^2=4mx

x=9/(4m)

A(9/(4m),-3）

|AF|=5

A到准线的距离为5

则A到y轴距离=5-|m|

|9/(4m)|=5-|m|

9=20|m|-4|m|^2

4|m|^2-20|m|+9=0

(2|m|-9)(2|m|-1)=0

|m|=9/2或|m|=1

m=±9/2或m=±1

抛物线线标准方程为：y^2=18x或y^2=-18x或y^2=4x或y^2=-4x.