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865 B
865 B
AF=A'F,ME=EM' 所以，问题就变成了求A'到M'的直线最短距离。
这里有一个天然的Rt\triangle A'AM'
OA=3,OM'=OM=3/2,所以AM'=9/2
下面求AA'是多长呢？我们知道A'是A关于抛物线的对称轴对称的，抛物线的对称轴是\large x=-\frac{b}{2a}=3,所以AA'=6
A'M'=\sqrt{(\frac{9}{2})^2+6^2}=\sqrt{\frac{81+144}{4}}=\frac{15}{2}=7.5