设A(x1,y1),B(x2,y2)两点再抛物线y=2x^2上,l是AB的垂直平分线
2个回答
1.若l经过焦点F,则|FA|=Y1+1/2P,|FB|=Y2+1/2P,又F是AB中垂线L上一点,有|FA|=|FB|,所以,Y1=Y2,则AB平行于X轴,根据抛物线关于Y轴D对称,则X1=-X2 ,所以X1+X2=0
2.设l与y轴交与点c(0,b),只要求b的范围即可,
l的斜率为2,则AB的斜率为-1/2
即（y2-y1)/(x2-x1)=-1/2
2(x2^2-x1^2)/(x2-x1)=-1/2
得 x1+x2=-1/4
∵|CA|=|CB|
∴ x1^2+(y1-b)^2=x2^2+（y2-b)^2
x1^2+y1^2-2by1=x2^2+y2^2-2by2
b=(x2^2-x1^2+y2^2-y1^2)/2(y2-y1)[将x1^2=1/2y1,x2^2=1/2y2带入]
=(1/2y2-1/2y1+y2^2-y1^2)/2(y2-y1)
=1/4+1/2(y1+y2)
=1/4+x1^2+x2^2
∵x1^2+x2^2＞2x1x2(x1≠x2)
∴2（x1^2+x2^2）＞(x1+x2)^2
∴x1^2+x2^2＞[(x1+x2)^2]/2
b＞1/4+[（x1+x2)^2/]2
=1/4+1/32
=9/32
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