高中数学 第二章 圆锥曲线 2_1 椭圆复习学案 新人教A版选修1-1

4.【答案】C

?a2?b2?1,22?xy22

△【解析】据题意可得?解得a=4,b=3,故椭圆方程为??1,故选C. 2b243?3,?|AB|?a?

5.【答案】D

6.【答案】

(Ⅰ) 设P(x,y),由椭圆定义可知,点P的轨迹C是以(0,?3),(0,3)为焦点,长半轴为2的椭圆.它的短半轴

y2?1. b?2?(3)?1,故曲线C的方程为x?4222?2y2?1,?x?(Ⅱ) 设A(x1,y1),B(x2,y2), 其坐标满足? 4?y?kx?1.?消去y并整理得(k2＋4)x2＋2kx－3=0, 故x1?x2??2k3xx??,. 12k2?4k2?4若OA?OB,即x1x2＋y1y2=0. 而y1y2=k2x1x2＋k(x1＋x2)＋1,

332k2于是x1x2?y1y2??2???1?0,

k?4k2?4k2?4化简得－4k2＋1=0,所以k??1. 2222(Ⅲ) |OA|2?|OB|2?x1?y12?(x2?y2)

22?(x12?x2)?4(1?x12?1?x2)

=－3(x1－x2)(x1＋x2)

?6k(x1?x2).

k2?43知x2<0, k2?4因为A在第一象限,故x1>0. 由x1x2??从而x1－x2>0.又k>0, 故|OA|2?|OB|2?0,

即在题设条件下,恒有|OA|?|OB|.

7.【答案】

?2b2?1?a?2?x2??(Ⅰ) 由条件得?a,所以方程?y2?1,

4?b?1?2b?a?(Ⅱ) 易知直线l斜率存在,令l:y=k(x+1),A(x1,y1),B(x2,y2),E(－4,y0),

?y?k(x?1)?22222

?由?x2(1+4k)x+8kx+4k－4=0,Δ=48k+16>0, 2??y?1?48k24k2?4,x1x2?, x1?x2??221?4k1?4kx1?1??(x1?1)??(x2?1)???由AQ??QB?(－1－x1,－y1)=λ(x2+1,y2)即?, 得,

x?1y???y2?12由AE??EB?(－4－x1,y0－y1)=μ(x2+4,y2－y0)即?x1?4??(x1?4)??(x1?4)， 得???,

x?42?y0?y1??(y2?y0)∴?????(x1?1)(x2?4)?(x1?4)(x2?1)2xx?5(x1?x2)?8??12

(x2?1)(x2?4)(x2?1)(x2?4)8k24k2?4将x1?x2??,x1x2?代入

1?4k21?4k28k2?840k28k2?8?40k2?8?32k2??8222有∴1?4k1?4k1?4k????????0.

(x2?1)(x2?4)(x2?1)(x2?4)