本文用初等方法讨论了与抛物线有关的若干几何最值问题,得到了十个有趣的结论.为方便读者摘用, 现用定理形式叙述如下:
定理1.抛物线的所有焦半径中,以过顶点的焦半径为最短.
证明:不妨设抛物线的极坐标方程为 ρ=                   ,则显然有ρ≥    ,其中等号成立当且仅当θ=2kπ+π (k∈Z)即焦半径通过抛物线的顶点时.证毕.
定理2.抛物线的过焦点的所有弦中,以抛物线的通径为最短.
证明:设抛物线极坐标方程为 ρ=                 ,焦点弦为AB,且设A(ρ1,θ),B(ρ2,θ+π),则有
         │AB│=ρ1+ρ2 =                 +                             =                       ≥ 2p =通径长,
其中等号成立当且仅当θ=kπ+π/2 (k∈Z) 即弦AB为通径时.证毕.
定理3.设A(a,0)是抛物线 y2=2px(p＞0)的对称轴上的定点,M(x,y)是抛物线上的动点,则

          │MA│m in =                                              

证明:由│MA│2= (x-a)2+y2=(x-a)2+2px = x2-2(a-p)x+a2  = [x-(a-p)]2+p(2a-p),并且注意到 x∈[0,+∞),立知结论成立.证毕.
定理4.设A(a,b)是抛物线 y2=2px(p＞0)内一定点,
F是焦点,M是抛物线上的动点,则                                                 y
          (│MA│+│MF│)min =a+p/2.                                          Q           M             A(a,b)
证明:如图1所示,作AQ⊥准线L:x=-p/2于Q,则知                             O  F                 x
         (│MA│+│MF│)min =│AQ│
        = a-(-p/2)=a+p/2.证毕.                                                                  图1
定理5.设线段AB是抛物线y2=2px(p＞0)的过焦点的弦,分别以A、B为切点的抛物线的两条切线相交于点M,则三角形ABM的面积的最小值为p2.
证明:设A(x1,y1),B(x2,y2),则由A、F、B三点共线可得:x1y2-x2y1=p/2·(y2-y1)……………(1)
于是利用(1)式由两切线方程                                                                 y                            
        AM: y1y=p(x+x1),                                                                                                    A         
        BM: y2y=p(x+x2),                                                                         M          F                 x   
易得M的坐标(x,y)适合 :                                                                              B      
                                                                                                                     
      ∵ 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                                                                                    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                   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                    X                                 t1+t2＝－(2ax0cosα-sinα)/a(cosα)2 ＝0………①                                    图6
t1t2＝(ax02-y0)/a(cosα)2 ＝－(m/2)2 ……………②
由①解出x0并代入②整理得
        y0＝      (secα)2＋       (cosα)2－       ……③
对③右边前两项利用基本不等式则得 y0≥2·     －      ＝（2ma-1)/4a. 于是,令
               (secα)2 ＝       (cosα)2,  得(cosα)2＝       .
因此, 当am≥1时,(y0)min＝（2ma-1)/4a ;
       当0＜am＜1时, 记(cosα)2＝x , 则③式化为关于x 的函数式
       y0＝f(x)＝    ·   ＋       ·x－        (0＜x≤1).
易证此函数是减函数, 故此时 (y0)min＝f(1)＝        .证毕.   
定理10. 设AB是抛物线 y2＝2px的焦点弦, O为坐标原点, 则三角形OAB的面积的最小值为 p2/2 .                                                                                                 y  
证明:(1)当AB⊥x轴时, 显然有 SΔAOB＝p2/2 ;                                          A
(2)当AB不垂直x轴时, 设AB: y＝k(x-p/2), 代                                 O        F                     x
入 y2＝2px并整理得 k2x2-(pk2+2p)x+k2p2/4=0. 于是                                    B
设A(x1,y1),B(x2,y2),则由弦长公式和韦达定理得:                                           图7
               │AB│＝   (1+k2 )[(x1+x2)2- 4x1x2]  
                            ＝                                                  ＝                   .
又顶点O到弦AB的距离
              d＝                   .
故此时 SΔAOB＝   │AB│·d＝     ·               ·                    
                      ＝    ·             ＞       .
综合(1)、(2), 定理10获证 .  

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