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春芽
共回答了15个问题采纳率:66.7% 举报
设P(m,n),A(x1,y1),B(x2,y2),
则对
x2
4+y2=1两边求导,得,[x/2+2yy′=0,
则过切点A的斜率为-
x1
4y1],切线方程为:y-y1=-
x1
4y1(x-x1),
又x12+4y12=4,化简即得PA:
x1x
4+y1y=1,
同理可得,PB:
x2x
4+y2y=1,
∵过P点作椭圆的切线PA,PB,
∴直线AB的方程为[mx/4]+ny=1.
代入椭圆方程可得(4n2+m2)x2-8mx+(16-16n2)=0,
∴x1+x2=[8m
4n2+m2,x1x2=
16?16n2
4n2+m2,
∴
/PA?
PB]=x1x2+m2-m(x1+x2)+y1y2-n(y1+y2)+n2
=x1x2+m2-m(x1+x2)+
(4?mx1)(4?mx2)
16n2-
8?m(x1+x2)
4+n2
=
20?3m2
4n2+m2+m2+n2-6,
∵m2+n2=16,
∴
PA?
PB=11-[44
3n2+16,
则当n=0,m=±4时,即P(±4,0),
/PA?
PB]有最小值[33/4].
故答案为:[33/4].
1年前
10