ll何处觅
幼苗
共回答了21个问题采纳率:81% 举报
(1)见解析(2)椭圆方程为
+y
2 =1.P点坐标为
(1)证明:设椭圆方程为
=1(a>b>0)①,则A(0,b),B(0,-b),T
.
AT:
=1②,BF:
=1③,解得交点C
,代入①得
=
=1,满足①式,则C点在椭圆上,即A、C、T
三点共线.
(2)过C作CE⊥x轴,垂足为E,则△OBF∽△ECF.
∵
=3
,CE=
b,EF=
c,则C
,代入①得
=1,∴a
2 =2c
2 ,b
2 =c
2 .设P(x
0 ,y
0 ),则x
0 +2
=2c
2 .此时C
,AC=
c,S
△ABC =
·2c·
=
c
2 ,
直线AC的方程为x+2y-2c=0,P到直线AC的距离为d=
,
S
△APC =
d·AC=
·
·
c=
·c.只须求x
0 +2y
0 的最大值,
(解法1)∵(x
0 +2y
0 )
2 =
+4
+2·2x
0 y
0 ≤
+4
+2(
+
)=3(
+2
)=6c
2 ,∴x
0 +2y
0 ≤
c.当且仅当x
0 =y
0 =
c时,(x
0 +2y
0 )
max =
c.
(解法2)令x
0 +2y
0 =t,代入
+2
=2c
2 得(t-2y
0 )
2 +2
-2c
2 =0,即6
-4ty
0 +t
2 -2c
2 =0.Δ=(-4t)
2 -24(t
2 -2c
2 )≥0,得t≤
c.当t=
c,代入原
1年前
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