Question

設(shè)橢圓=1(a＞b＞0)的焦點分別為F₁(-1,0)、F₂(1,0),右準線l交x軸于點A,且.

(1)試求橢圓的方程;

(2)過F₁、F₂分別作互相垂直的兩直線與橢圓分別交于D、E、M、N四點(如圖所示),試求四邊形DMEN面積的最大值和最小值.

(文)已知函數(shù)f(x)=x³+bx²+cx,b、c∈R,且函數(shù)f(x)在區(qū)間(-1,1)上單調(diào)遞增,在區(qū)間(1,3)上單調(diào)遞減.

(1)若b=-2,求c的值;

(2)求證:c≥3;

(3)設(shè)函數(shù)g(x)=f′(x),當(dāng)x∈［-1,3］時,g(x)的最小值是-1,求b、c的值.

Answer 1

解:(1)由題意,||=2c=2,

∴A(a²,0).                                                              

∵,

∴F₂為AF₁的中點.                                                      

∴a²=3,b²=2,

即橢圓方程為=1.                                                 

(2)當(dāng)直線DE與x軸垂直時,|DE|=2,

此時|MN|=2a=,四邊形DMEN的面積為=4.

同理當(dāng)MN與x軸垂直時,也有四邊形DMEN的面積為=4.         

當(dāng)直線DE、MN均與x軸不垂直時,

設(shè)DE:y=k(x+1),

代入橢圓方程,消去y得

(2+3k²)x²+6k²x+(3k²-6)=0.

設(shè)D(x₁,y₁),E(x₂,y₂),

則                                                   

∴|x₁-x₂|=.

∴|DE|=|x₁-x₂|=.

同理,|MN|=.                               

∴四邊形的面積

S=.        

令u=k²+,得S=,

∵u=k²+≥2,

當(dāng)k=±1時,u=2,S=,且S是以u為自變量的增函數(shù),

∴≤S＜4.

綜上,可知四邊形DMEN面積的最大值為4,最小值為.                     

(文)解:(1)由已知可得f′(1)=0,                                            

又f′(x)=x²+2bx+c,

∴f′(1)=1+2b+c=0.                                                       

將b=-2代入,可得c=3.                                                    

(2)證明:由(1)可知b=,代入f′(x)可得f′(x)=x²-(c+1)x+c.

令f′(x)=0,則x₁=1,x₂=c,                                                   

又當(dāng)-1＜x＜1時,f′(x)≥0;

當(dāng)1＜x＜3時,?f′(x)≤0.

如圖所示.

易知c≥3.                                                               

(3)若1≤-b≤3,則

g(x)_min=g(-b)=b²-2b²+c=-1.

又1+2b+c=0,得b=-2或b=0(舍),c=3.

若-b≥3,則g(x)_min=g(3)=9+6b+c=-1,

又1+2b+c=0,得b=(舍).

綜上所述,b=-2,c=3.