梦血 幼苗
共回答了18个问题采纳率:88.9% 举报
p |
2 |
p |
2 |
(1)由已知得焦点F(0,
p
2),准线方程为y=−
p
2,
由抛物线的定义及点N的纵坐标为1,得|NF|=
p
2+yN=
p
2+1
又|NF|=2,
∴
p
2+1=2∴p=2,
∴抛物线的方程为x2=4y(4分)
(2)依题意设直线l的方程为:y=kx+1(k必存在)
y=kx+1
x2=4y⇒x2−4kx−4=0,…(6分)
则△=16k2+16>0,
设直线l与抛物线的交点坐标为A(x1,y1),B(x2,y2),
则有x1+x2=4k,x1x2=-4,…(8分)
∴|AB|=
1+k2|x1−x2|=
1+k2•
(x1+x2)2−4x1x2=4(1+k2)…(10分)
∵O到AB的距离d=
1
点评:
本题考点: 直线与圆锥曲线的综合问题.
考点点评: 本题考查抛物线的定义与方程,考查直线与抛物线的位置关系,考查弦长的计算,考查三角形面积的计算,正确运用韦达定理是关键.
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