C于P、Q 两点,且M为线段PQ的中点,OP⊥OQ.求直线l的方程及椭圆 C的方程. 解析(1)由|AB| = 52 |BF|, 得 a +b = 52 a,即4a +4b =5a, ∴4a +4(a -c) =5a,∴e = ca = 32 . (2)由(1)知椭圆C: x4b + yb =1. 设P(x,y),Q(x,y),∵M - 16 17 , 217 ( ) 是线段PQ的中点, ∴x +x =- 32 17 ,y +y = 417 , 由 x4b + yb =1, x4b + yb =1,得 x -x 4b + y -y b =0, 即(x +x)(x -x) 4b + (y +y)(y -y) b =0, ∴- 32 17 (x -x) 4 + 417 (y -y) =0,从而k = y -y x -x =2, 进而可得直线l的方程为y - 217 =2x -- 16 17() [ ] ,即2x -y +2 =0, 由 2x -y +2 =0, x4b + yb =1 { ?x +4(2x +2) -4b =0, 即17x +32x +16 -4b =0. Δ=32 +16 ×17(b -4)>0?b> 2 17 17 . x +x =- 32 17 ,xx = 16 -4b 17 . ∵OP⊥OQ,∴OP →·OQ →=0, 即xx +yy =0,xx +(2x +2)(2x +2) =0,5xx +4(x + x) +4 =0. 从而 5(16 -4b) 17 - 128 17 +4 =0,解得b =1.满足题意. ∴椭圆C的方程为 x4 +y =1. ????????????????????????????????????????????????????????????????????????????????????????????????????????
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