题目内容
抛物线y2=4x的焦点为F,过点F的直线交抛物线于A,B两点,且AF=2BF,则A点的坐标为______.
设A(x1,y1),B(x2,y2)
设过点F的直线方程为x=my+1,代入抛物线y2=4x,得y2-4my-4=0
∵△>0,∴y1+y2=4m,y1•y2=-4
∵AF=2BF,∴y1=-2y2
∴y1=2
,m=
或y1=-2
,m=-
,代入x=my+1得
∴x1=5,y1=2
,或x1=5,y1=-2
,
∴A点的坐标为(5,2
)或(5,-2
)
故答案为(5,2
)或(5,-2
)
设过点F的直线方程为x=my+1,代入抛物线y2=4x,得y2-4my-4=0
∵△>0,∴y1+y2=4m,y1•y2=-4
∵AF=2BF,∴y1=-2y2
∴y1=2
| 2 |
| 2 |
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∴x1=5,y1=2
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| 2 |
∴A点的坐标为(5,2
| 2 |
| 2 |
故答案为(5,2
| 2 |
| 2 |
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