Question

A为抛物线x^2=4y上异于原点的任意一点，F为抛物线焦点，l为抛物线在A点处的切线

A为抛物线x^2=4y上异于原点的任意一点，F为抛物线焦点，l为抛物线在A点处的切线，点BC在抛物线上，AB⊥l且交y轴于M，点AFC共线，直线BC交y轴于N. （1）求证|AF|=|MF| （2）求|MN|的最小值

Answer 1

设点A坐标为(a,a²/4)
4y=x²对x求导得：y'=x/2
所以直线I斜率为a/2,直线AB斜率为-2/a
AB直线方程为y-a²/4=(-2/a)(x-a),令x=0解得M点坐标(0,2+a²/4),与抛物线方程联立解得B点坐标(-a-8/a,4+16/a² +a²/4)
F点坐标(0,1),AF斜率=(a²/4 -1)/(a-0)=a/4 -1/a, AF直线方程y-a²/4=(a/4-1/a)(x-a),与抛物线联立解得C点坐标(-4/a,4/a²）求出BC方程，与y轴交点N点坐标(0,-1-8/a²）
|AF|=A到准线y=-1的距离=1+a²/4,|MF|=2+a²/4 -1=1+a²/4
∴|AF|=|MF|
|MN|=2+a²/4+1+8/a²=3+a²/4 +8/a²≥3+2√2

Answer 2

【1】
略
【2】
|MN|min=3+2√2

Answer 3

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