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顶点在原点,焦点在X轴上的抛物线被直线y=-2x-1所得弦长AB等于5根号3,求抛物线方程?
人气:347 ℃ 时间:2019-11-06 18:17:54
解答

焦点(-p/2,0) ,
设抛物线方程为 :
y^2=-2px(p>0)
将直线代入
(-2x-1)^2=4x^2+4x+1=-2px
4x^2+(4+2p)x+1=0
x1+x2=-(4+2p)/4 ,
x1x2=1/4
又因为|AB|^2=(x1-x2)^2+(y1-y2)^2=(5√3)^2
=(x1-x2)^2+4(x1-x2)^2
=5(x1-x2)^2
=5[(x1+x2)^2-4x1x2]
=5[(-(4+2p)/4)^2-4*1/4]
=5[(p+2)^2-1]
=25*3
=> (p+2)^2-1=15解得p=2(p=-6舍弃)
∴抛物线方程为:y^2=-2px=-4x
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