【答案】
分析:(1)因?yàn)橹本€過橢圓的左頂點(diǎn)與上頂點(diǎn),故可解出直線與坐標(biāo)軸的交點(diǎn),即知橢圓的長(zhǎng)半軸長(zhǎng)與短半軸長(zhǎng),依定義寫出橢圓的方程即可.
(2)法一、引入直線AS的斜率k,用點(diǎn)斜式寫出直線AS的方程,與l的方程聯(lián)立求出點(diǎn)M的坐標(biāo),以及點(diǎn)S的坐標(biāo),又點(diǎn)B的坐標(biāo)已知,故可解 出直線SB的方程,亦用參數(shù)k表示的方程,使其與直線l聯(lián)立,求出點(diǎn)N的坐標(biāo),故線段MN的長(zhǎng)度可以表示成直線AS的斜率k的函數(shù),根據(jù)其形式選擇單調(diào)性法或者基本不等式法求最值,本題適合用基本不等式求最值.
法二、根據(jù)圖形構(gòu)造出了可用基本不等式的形式來求最值.
(3)在上一問的基礎(chǔ)上求出參數(shù)k,則直線SB的方程已知,可求出線段AB的長(zhǎng)度,若使面積為

,只須點(diǎn)T到直線BS的距離為

即可,由此問題轉(zhuǎn)化為研究與直線SB平行且距離為

的直線與橢圓的交點(diǎn)個(gè)數(shù)問題,下易證
解答:
解:(1)由已知得,橢圓C的左頂點(diǎn)為A(-2,0),
上頂點(diǎn)為D(0,1),∴a=2,b=1
故橢圓C的方程為

(4分)
(2)依題意,直線AS的斜率k存在,且k>0,故可設(shè)直線AS的方程為y=k(x+2),從而

,由

得(1+4k
2)x
2+16k
2x+16k
2-4=0
設(shè)S(x
1,y
1),則

得

,從而

即
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,(6分)
又B(2,0)由

得

,
∴
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,(8分)
故

又k>0,∴

當(dāng)且僅當(dāng)
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,即

時(shí)等號(hào)成立.
∴

時(shí),線段MN的長(zhǎng)度取最小值

(10分)
(2)另解:設(shè)S(x
s,y
S),

依題意,A,S,M三點(diǎn)共線,且所在直線斜率存在,
由k
AM=k
AS,可得

同理可得:

又

所以,

=

不仿設(shè)y
M>0,y
N<0

當(dāng)且僅當(dāng)y
M=-y
N時(shí)取等號(hào),
即

時(shí),線段MN的長(zhǎng)度取最小值

.
(3)由(2)可知,當(dāng)MN取最小值時(shí),

此時(shí)BS的方程為

,∴

(11分)
要使橢圓C上存在點(diǎn)T,使得△TSB的面積等于

,只須T到直線BS的距離等于

,
所以T在平行于BS且與BS距離等于

的直線l'上.
設(shè)直線l':x+y+t=0,則由

,解得

或

.
又因?yàn)門為直線l'與橢圓C的交點(diǎn),所以經(jīng)檢驗(yàn)得

,此時(shí)點(diǎn)T有兩個(gè)滿足條件.(14分)
點(diǎn)評(píng):本題是解析幾何中直線與圓錐曲線位置關(guān)系中很復(fù)雜的題目,要求答題者擁有較高的探究轉(zhuǎn)化能力以及對(duì)直線與圓錐曲線位置關(guān)系中特征有較好 的理解,且符號(hào)運(yùn)算能力較強(qiáng)才能勝任此類題的解題工作,這是一個(gè)能力型的題,好題.