关于抛物线的十个最值问题

;                                                          ∵ kMF·kAF=-1, ∴MF⊥AB,即│MF│是△MAB的AB边上的高.          图2  ∵ │MF│≥│FK│(焦点F到准线x=-p/2的距离)=p,  又由定理2知│AB│≥2p(通径长),  ∴ S△MAB=1/2·│AB│·│MF│≥1/2·2p·p=p2,  因其中等号当且仅当AB⊥x轴时成立,故三角形MAB的最小值为p2.证毕.  定理6.过抛物线y2=2px的顶点O引两条互相垂直的动弦OA和OB,则三角形OAB的面积的最小值为4p2.                                                                                                      y  证明:设A(x1,y1),B(x2,y2),则由OA⊥OB得                                                                     A             x1x2+y1y2=0 ……………………………………(1)                                O                            x  将y12=2px1, y22=2px2代入(1)立得: x1x2=4p2…………(2)                                               于是                                                                                                                       B             (S△OAB)2 =1/4·│OA│2·│OB│2                      &nb

sp;                                                             图3                   =1/4·(x12+y12)·(x22+y22) =1/4·(x12+2px1)·(x22+2px2)                                     =1/4·[(x1x2)2+2px1x2 (x1+x2)+4p2x1x2]                  ≥1/4·[(x1x2)2+2px1x2 (2√x1x2)+4p2x1x2]………………………………………(3)  将(2)式代入(3)则得 (S△OAB)2≥16p4,从而S△OAB≥4p2,因其中等号当x1=x2=2p时取到,故三角形OAB的面积的最小值为4p2。证毕.  定理7.抛物线 y2=2px的内接等腰直角三角形的面积的最小值为4p2.  证明:设Rt△ABC内接于抛物线 y2=2px,点C为直角顶点,设A(x1,y1),B(x2,y2),C(x3,y3),根据抛物线的对称性以及其开口方向,不妨设 y1＞0,y2＜y3≤0,并记直线CA的斜率为k,则由         y3-y1=k(x3-x1)=k(y32/2p －y12/2p) 及                                              y         y3-y2=-1/k·(x3-x2)=-1/k·(y32/2p－y22/2p)                                          A  可得       y1 =2p/k－y3 及 y2=-2pk-y3………………(1)                                O                         x  又由      │AC│=│BC│有                                                                        C              

《关于抛物线的十个最值问题(第2页)》