(Ⅱ)解法1:依题意,可设直线l的方程为y=kx+2,代入双曲线C的方程并整理得(1-K)x2-4kx-6=0.
∵直线l与双曲线C相交于不同的两点E、F,
2???k??1?1-k?0∴ ?? ?22????3?k????(?4k)?4?6(1?k)?02
3∴k∈(-3,-1)∪(-1,1)∪(1,3). 设E(x,y),F(x2,y2),则由①式得x1+x2=|EF|=(x1?x2)2?(y1?x2)2?=1?k24k1?k22,x1x2??261?k,于是
(1?k)(x1?x2)
1?k2?(x1?x2)?4x1x2?2?223?k221?k.
而原点O到直线l的距离d=
21?k2,
∴S△DEF=
12d?EF?12?21?k2?1?k2?223?k221?k?223?k1?k22.
若△OEF面积不小于22,即S△OEF?22,则有
223?k221?k?22?k4?k2?2?0,解得?2?k?2. ③
综合②、③知,直线l的斜率的取值范围为[-2,-1]∪(1-,1) ∪(1,
2).
解法2:依题意,可设直线l的方程为y=kx+2,代入双曲线C的方程并整理, 得(1-K2)x2-4kx-6=0.
∵直线l与双曲线C相交于不同的两点E、F,
2???1-k?0?k??1? ?∴ ?22????3?k????(?4k)?4?6(1?k)?0
3.∴k∈(-3,-1)∪(-1,1)∪(1,3). 设E(x1,y1),F(x2,y2),则由①式得
|x1-x2|=(x1?x2)?4x1x2?2?1?k2?223?k221?k. ③
当E、F在同一去上时(如图1所示), S△OEF=S?ODF?S?ODE?12OD?x1?x2?12OD?x1?x2;
当E、F在不同支上时(如图2所示).
S?OEF?S?ODF?S△ODE=
12OD?(x1?x2)?12OD?x1?x2.
综上得S△OEF=
12OD?x1?x2,于是
由|OD|=2及③式,得S△OEF=
223?k221?k.
若△OEF面积不小于22,即S?OEF?22,则有
223?k221?k?22?k4?k2?0,解得?2?k?2. ④
综合②、④知,直线l的斜率的取值范围为[-2,-1]∪(-1,1)∪(1,2).
18. (Ⅰ)设OA?m?d,AB?m,OB?m?d
由勾股定理可得:(m?d)?m?(m?d) 得:d?14m,tan?AOF?ba222,tan?AOB?tan2?AOF?ABOA?43
2ba2由倍角公式??43,解得
ba?12?b?1????a?,则离心率e?52.
(Ⅱ)过F直线方程为y??ab(x?c),与双曲线方程
xa22?yb22?1联立
将a?2b,c?5b代入,化简有
154b2x?285bx?21?0
4??a?1????b?2x1?x2?2?a???2 (x?x)?4x1x2??1?????12???b?????将数值代入,有4???325b5??15????x22?28b?4??5?2??,解得b?3 ??故所求的双曲线方程为
36?y29?1。
19. 解:由条件知F(?2,0),F12(2,0),设A(x1,y1),B(x2,y2).
?????????(I)解法一:(I)设M(x,y),则则F1M?(x?2,y),F1A?(x1?2,y1), ?????????????????????????F1B?(x2?2,y2),F1O?(2,0),由F1M?F1A?F1B?F1O得 ?x?2?x1?x2?6,?x1?x2?x?4,即? ??y?y1?y2?y1?y2?y于是AB的中点坐标为??x?4y,22???. ?y当AB不与x轴垂直时,
y1?y2x1?x2?2x?42??2yx?8,即y1?y2?yx?8(x1?x2).
2222又因为A,B两点在双曲线上,所以x1?y1?2,x2?y2?2,两式相减得
(x1?x2)(x1?x2)?(y1?y2)(y1?y2),即(x1?x2)(x?4)?(y1?y2)y.
将y1?y2?yx?8(x1?x2)代入上式,化简得(x?6)?y?4.
22当AB与x轴垂直时,x1?x2?2,求得M(8,0),也满足上述方程. 所以点M的轨迹方程是(x?6)?y?4. ?x1?x2?x?4,解法二:同解法一的(I)有?
?y1?y2?y22当AB不与x轴垂直时,设直线AB的方程是y?k(x?2)(k??1). 代入x?y?2有(1?k)x?4kx?(4k?2)?0.
222222
则x1,x2是上述方程的两个实根,所以x1?x2?4k22k?1.
?4k2?4k. y1?y2?k(x1?x2?4)?k??4??2?k?1?k?1由①②③得x?4?4kk?124k22k?1.…………………………………………………④
y?.……………………………………………………………………⑤
x?4y当k?0时,y?0,由④⑤得,?k,将其代入⑤有
4?y?x?4y22(x?4)y??14y(x?4)(x?4)?y22.整理得(x?6)2?y2?4.
当k?0时,点M的坐标为(4,0),满足上述方程.
当AB与x轴垂直时,x1?x2?2,求得M(8,0),也满足上述方程. 故点M的轨迹方程是(x?6)2?y2?4.
????????(II)假设在x轴上存在定点C(m,0),使CA?CB为常数.
当AB不与x轴垂直时,设直线AB的方程是y?k(x?2)(k??1). 代入x?y?2有(1?k)x?4kx?(4k?2)?0.
4k22222222则x1,x2是上述方程的两个实根,所以x1?x2?k?1,x1x2?4k?2k?122,
????????2于是CA?CB?(x1?m)(x2?m)?k(x1?2)(x2?2)
?(k?1)x1x2?(2k?m)(x1?x2)?4k?m
2222?(k?1)(4k?2)k?12(1?2m)k?2k?122222?4k(2k?m)k?1222?4k?m
22??m?2(1?2m)?24?4mk?12?m.
2
????????????????因为CA?CB是与k无关的常数,所以4?4m?0,即m?1,此时CA?CB=?1.
当AB与x轴垂直时,点A,B的坐标可分别设为(2,2),(2,?????????此时CA?CB?(1,2)?(1,?2)??1.
2),
????????故在x轴上存在定点C(1,0),使CA?CB为常数.
20.(Ⅰ)由题意知,双曲线C的顶点(0,a)到渐近线ax?by?0的距离为255,
所以
aba?b22?255所以abc?255
?ab25??c5??a?2??5?c得?b?1 由??2?a?222?c?5?c?a?b???所以曲线C的方程是
y24?x?1
2(Ⅱ)设直线AB的方程为y?kx?m, 由题意知k?2,m?0
?y?kx?m?y?2xm2m由?得A点的坐标为(2?k2?k?m2m,),
由??y?kx?m?y??2x得B点的坐标为(2?k2?k,),
uuuruurm1?2m1?AP??PB,得P点的坐标为((?),(?)
1??2?k2?k1??2?k2?k将P点的坐标代入
y24?x?1得
24m224?k?(1??)2?
设Q为直线AB与y轴的交点,则Q点的坐标为(0,m)
S?AOB=S?AOQ?S?BOQ
???121212OQgxA?m(m2?k1?12OQgxB?)?112m(xA?xB)2m2?kg224?k4m
(???)?1
双曲线习题及答案
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