關于拋物線的十個最值問題

本文用初等方法討論了與拋物線有關的若干幾何最值問題,得到了十個有趣的結論.為方便讀者摘用, 現(xiàn)用定理形式敘述如下:  定理1.拋物線的所有焦半徑中,以過頂點的焦半徑為最短.  證明:不妨設拋物線的極坐標方程為 ρ=                   ,則顯然有ρ≥    ,其中等號成立當且僅當θ=2kπ+π (k∈Z)即焦半徑通過拋物線的頂點時.證畢.  定理2.拋物線的過焦點的所有弦中,以拋物線的通徑為最短.  證明:設拋物線極坐標方程為 ρ=                 ,焦點弦為AB,且設A(ρ1,θ),B(ρ2,θ+π),則有           │AB│=ρ1+ρ2 =                 +                             =                       ≥ 2p =通徑長,  其中等號成立當且僅當θ=kπ+π/2 (k∈Z) 即弦AB為通徑時.證畢.  定理3.設A(a,0)是拋物線 y2=2px(p＞0)的對稱軸上的定點,M(x,y)是拋物線上的動點,則              │MA│m in =                                                  證明: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,+∞),立知結論成立.證畢.  定理4.設A(a,b)是拋物線 y2=2px(p＞0)內(nèi)一定點,  F是焦點,M是拋物線上的動點,則                                                 y            (│MA│+│MF│)min =a+p/2.                                          Q           M             A(a,b)  證明:如圖1所示,作AQ⊥準線L:x=-p/2于Q,則知                             O  F                 x           (│MA│+│MF│)min =│AQ│          = a-(-p/2)=a+p/2.證畢.                                                                  圖1  定理5.設線段AB是拋物線y2=2px(p＞0)的過焦點的弦,分別以A、B為切點的拋物線的兩條切線相交于點M,則三角形ABM的面積的最小值為p2.  證明:設A(x1,y1),B(x2,y2),則由A、F、B三點共線可得:x1y2-x2y1=p/2·(y2-y1)……………(1)  于是利用(1)式由兩切線方程                                                                 y                                      AM: y1y=p(x+x1),                                                                                                    A                   BM: y2y=p(x+x2),                                                                         M          F                 x     易得M的坐標(x,y)適合 :                                                                              B                                                                                                                                     ∵ kMF·kAF=-1, ∴MF⊥AB,即│MF│是△MAB的AB邊上的高.          圖2  ∵ │MF│≥│FK│(焦點F到準線x=-p/2的距離)=p,  又由定理2知│AB│≥2p(通徑長),  ∴ S△MAB=1/2·│AB│·│MF│≥1/2·2p·p=p2,  因其中等號當且僅當AB⊥x軸時成立,故三角形MAB的最小值為p2.證畢.  定理6.過拋物線y2=2px的頂點O引兩條互相垂直的動弦OA和OB,則三角形OAB的面積的最小值為4p2.                                                                                                      y  證明:設A(x1,y1),B(x2,y2),則由OA⊥OB得                                                                     A             x1x2+y1y2=0 ……………………………………(1)                                O                            x  將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

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sp;                                                             圖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)  將(2)式代入(3)則得 (S△OAB)2≥16p4,從而S△OAB≥4p2,因其中等號當x1=x2=2p時取到,故三角形OAB的面積的最小值為4p2。證畢.  定理7.拋物線 y2=2px的內(nèi)接等腰直角三角形的面積的最小值為4p2.  證明:設Rt△ABC內(nèi)接于拋物線 y2=2px,點C為直角頂點,設A(x1,y1),B(x2,y2),C(x3,y3),根據(jù)拋物線的對稱性以及其開口方向,不妨設 y1＞0,y2＜y3≤0,并記直線CA的斜率為k,則由         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)                                      圖4  將x1=y12/2p,x2=y22/2p,x3=y32/2p及(1)代入(2)可得  y3=                    …………………………(3)  從而據(jù)(1)、(3)可得     y1-y3=             ………………………………………………………(4)    于是△ABC的面積              S=1/2·│AC│2 =1/2·[(x1-x3)2+(y1-y3)2]=    ·         ·(y1-y3)2                =    2p2  ·          ·(             )2                              =2p2·               ·            ≥2p2·         ·                       =4p2.    因當k=1且y3=0時上式等號成立,故等腰Rt△ABC面積的最小值為4p2.證畢.  定理8.設AB是拋物線的焦點弦, 準線與拋物線對稱軸的交點為M, 則∠AMB的最大值 為π/2.  證明:如圖5所示, 設A1、B1分別是A、B在準線L上的                              y                        射影, F是焦點, 連A1F和B1F, 則知                                                              A         A  (1)當AB⊥MF時, 顯然有∠AMB＝π/2;                                            M       F                 X  (2)當AB與MF不垂直時, 由│AA1│＞│A1M│知                           B1               B  ∠AMA1＞∠A1AM＝π/2－∠AMA1,                                                    圖5  ∴     ∠AMA1＞π/4;  同理  ∠BMB1＞π/4, 故有∠AMB＜π/2.  綜合(1)、(2), 定理8獲證.  定理9.設AB是拋物線 y＝a x2 (a＞0) 的長為定長m的動弦, 則  Ⅰ.當m≥1/a (通徑長)時, AB的中點M到x軸的距離的最小值為（2ma-1)/4a ;                Ⅱ.當m＜1/a (通徑長)時, AB的中點M到x軸的距離的最小值為 am2/4.  證明:設M(x0,y0), 將直線AB的參數(shù)方程                                                                                                                 y                             (其中t為參數(shù),傾斜角α≠π/2)                                       A           代入y＝ax2 并整理得                                                                              M   a(cosα)2·t2+(2ax0cosα-sinα)·t+(ax02-y0)＝0,                             B                        故由韋達定理和參數(shù) t的幾何意義以及│AB│＝m 立得                          0                    X                                 t1+t2＝－(2ax0cosα-sinα)/a(cosα)2 ＝0………①                                    圖6   t1t2＝(ax02-y0)/a(cosα)2 ＝－(m/2)2 ……………②  由①解出x0并代入②整理

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得          y0＝      (secα)2＋       (cosα)2－       ……③  對③右邊前兩項利用基本不等式則得 y0≥2·     －      ＝（2ma-1)/4a. 于是,令                 (secα)2 ＝       (cosα)2,  得(cosα)2＝       .  因此, 當am≥1時,(y0)min＝（2ma-1)/4a ;         當0＜am＜1時, 記(cosα)2＝x , 則③式化為關于x 的函數(shù)式         y0＝f(x)＝    ·   ＋       ·x－        (0＜x≤1).  易證此函數(shù)是減函數(shù), 故此時 (y0)min＝f(1)＝        .證畢.     定理10. 設AB是拋物線 y2＝2px的焦點弦, O為坐標原點, 則三角形OAB的面積的最小值為 p2/2 .                                                                                                 y    證明:(1)當AB⊥x軸時, 顯然有 SΔAOB＝p2/2 ;                                          A  (2)當AB不垂直x軸時, 設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  設A(x1,y1),B(x2,y2),則由弦長公式和韋達定理得:                                           圖7                 │AB│＝   (1+k2 )[(x1+x2)2- 4x1x2]                                ＝                                                  ＝                   .  又頂點O到弦AB的距離                d＝                   .  故此時 SΔAOB＝   │AB│·d＝     ·               ·                                            ＝    ·             ＞       .  綜合(1)、(2), 定理10獲證 .        

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