关于抛物线的十个最值问题

;     B                (x1-x3)2+(y1-y3)2=(x3-x2)2+(y3-y2)2 …………(2)                                      图4  将x1=y12/2p,x2=y22/2p,x3=y32/2p及(1)代入(2)可得  y3=                    …………………………(3)  从而据(1)、(3)可得     y1-y3=             ………………………………………………………(4)    于是△ABC的面积              S=1/2·│AC│2 =1/2·[(x1-x3)2+(y1-y3)2]=    ·         ·(y1-y3)2                =    2p2  ·          ·(             )2                              =2p2·               ·            ≥2p2·         ·                       =4p2.    因当k=1且y3=0时上式等号成立,故等腰Rt△ABC面积的最小值为4p2.证毕.  定理8.设AB是抛物线的焦点弦, 准线与抛物线对称轴的交点为M, 则∠AMB的最大值 为π/2.  证明:如图5所示, 设A1、B1分别是A、B在准线L上的                              y                        射影, F是焦点, 连A1F和B1F, 则知                                                              A         A  (1)当AB⊥MF时, 显然有∠AMB＝π/2;                                            M       F                 X  (2)当AB与MF不垂直时, 由│AA1│＞│A1M│知                           B1               B  ∠AMA1＞∠A1AM＝π/2－∠AMA1,                                                    图5  ∴     ∠AMA1＞π/4;  同理  ∠BMB1＞π/4, 故有∠AMB＜π/2.  综合(1)、(2), 定理8获证.  定理9.设AB是抛物线 y＝a x2 (a＞0) 的长为定长m的动弦, 则  Ⅰ.当m≥1/a (通径长)时, AB的中点M到x轴的距离的最小值为（2ma-1)/4a ;                Ⅱ.当m＜1/a (通径长)时, AB的中点M到x轴的距离的最小值为 am2/4.  证明:设M(x0,y0), 将直线AB的参数方程                                                                                                                 y                             (其中t为参数,倾斜角α≠π/2)                                       A           代入y＝ax2 并整理得                                                                              M   a(cosα)2·t2+(2ax0cosα-sinα)·t+(ax02-y0)＝0,                             B                        故由韦达定理和参数 t的几何意义以及│AB│＝m 立得                          0           &

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