Question

(理)已知F₁(-2,0),F₂(2,0),點(diǎn)P滿足|PF₁|-|PF₂|=2,記點(diǎn)P的軌跡為E.

(1)求軌跡E的方程;

(2)若直線l過點(diǎn)F₂且與軌跡E交于P、Q兩點(diǎn).

①無論直線l繞點(diǎn)F₂怎樣轉(zhuǎn)動(dòng),在x軸上總存在定點(diǎn)M(m,0),使MP⊥MQ恒成立,求實(shí)數(shù)m的值.

②過P、Q作直線x=的垂線PA、QB,垂足分別為A、B,記λ=,求λ的取值范圍.

(文)已知等差數(shù)列{a_n}中,a₁=-2,a₂=1.

(1)求{a_n}的通項(xiàng)公式;

(2)調(diào)整數(shù)列{a_n}的前三項(xiàng)a₁、a₂、a₃的順序,使它成為等比數(shù)列{b_n}的前三項(xiàng),求{b_n}的前n項(xiàng)和.

Answer 1

答案：(理)解:(1)由|PF₁|-|PF₂|=2＜|F₁F₂|,知點(diǎn)P的軌跡E是以F₁、F₂為焦點(diǎn)的雙曲線右支,由c=2,2a=2,∴b²=3.故軌跡E的方程為x²=1(x≥1).

(2)當(dāng)直線l的斜率存在時(shí),設(shè)直線方程為y=k(x-2),P(x₁,y₁),Q(x₂,y₂),與雙曲線方程聯(lián)立消去y,得(k²-3)x²-4k²x+4k²+3=0.

∴

①∵=(x₁-m)(x₂-m)+y₁y₂=(x₁-m)(x₂-m)+k²(x₁-2)(x₂-2)

=(k²+1)x₁x₂-(2k²+m)(x₁+x₂)+m²+4k²

==+m².

∵M(jìn)P⊥MQ,∴=0.故得3(1-m²)+k²(m²-4m-5)=0對(duì)任意的k²＞3恒成立,

∴解得m=-1.∴當(dāng)m=-1時(shí),MP⊥MQ.

當(dāng)直線l的斜率不存在時(shí),由P(2,3),Q(2,-3)及M(-1,0),知結(jié)論也成立,綜上,當(dāng)m=-1時(shí),MP⊥MQ.

②∵a=1,c=2,

∴直線x=是雙曲線的右準(zhǔn)線.由雙曲線定義,得|PA|=|PF₂|=|PF₂|,|QB|=|QF₂|.

∴λ==.

∵k²＞3,∴0＜＜,故＜λ＜.

注意到直線的斜率不存在時(shí),|PQ|=|AB|,此時(shí)λ=.綜上,λ=[,).

(文)解：(1)由已知,得a₂-a₁=1-(-2)=3,∴{a_n}的公差d=3.∴a_n=a₁+(n-1)d=-2+3(n-1)=3n-5.

(2)由(1)得a₃=a₂+d=1+3=4,∴a₁=-2,a₂=1,a₃=4.依題意,可得數(shù)列{b_n}的前三項(xiàng)為b₁=1,b₂=-2,b₃=4或b₁=4,b₂=-2,b₃=1.

①當(dāng)數(shù)列{b_n}的前三項(xiàng)為b₁=1,b₂=-2,b₃=4時(shí),則q=-2,

∴S_n==[1-(-2)ⁿ].

②當(dāng)數(shù)列{b_n}的前三項(xiàng)為b₁=4,b₂=-2,b₃=1時(shí),則q=-.

∴S_n==[1-(-)ⁿ].