Question

已知顶点在原点，焦点在x轴上的抛物线被直线y=2x+1截得的弦长为√15，求抛物线的方程

Answer 1

假设y^=2px,把y=2x+1带入，
4x^-(4+2p)x+1=0,
x1+x2=(4+2p)/4,
x1x2=1/4,
｜x1-x2｜=√〔(4+2p)^/16 -1〕,
弦长^2=15=（1+k^)｜x1-x2｜=5〔(4+2p)^/16 -1〕,
解得：p=2, -6;抛物线的方程:y^=4x 或者y^=-12x。

Answer 2

设y^2=2px
代入直线得：4x^2+4x+1=2px
x^2+(1-p/2)x+1/4=0
弦长的平方=(x1-x2)^2+(y1-y2)^2=(x1-x2)^2+(2x1-2x2)^2
=5[(x1+x2)^2-4x1x2]
=5[ (1-p/2)^2-1]=15
解得p=-2, 6
方程 y^2=-4x或y^2=12x

Answer 3

设y^2=ax 与y=2x+1联立得4x^2+(4-a)x+1=0 x1+x2=a/4-1 x1x2=1/4
15=(1+4)[(a/4-1)^2-1] a=12 a=-4
y^2=12x或y^2=-4x