设P(x,y),Q(x′,y′)是椭圆x2a2+y2b2=1(a>0,b>0)上的两点,则下列四个结论:①a2+b2≥(x+y)2;②1x2+1y2≥(1a+1b)2;③a2x2+b2y2≥4;④xx

题目简介

设P(x,y),Q(x′,y′)是椭圆x2a2+y2b2=1(a>0,b>0)上的两点,则下列四个结论:①a2+b2≥(x+y)2;②1x2+1y2≥(1a+1b)2;③a2x2+b2y2≥4;④xx

题目详情

设 P(x,y),Q(x′,y′) 是椭圆 
x2
a2
+
y2
b2
=1
(a>0,b>0)上的两点,则下列四个结论:①a2+b2≥(x+y)2;②
1
x2
+
1
y2
≥(
1
a
+
1
b
)2
;③
a2
x2
+
b2
y2
≥4
;④
xx′
a2
+
yy′
b2
≤1
.其中正确的个数为(  )
A.1个B.2个C.3个D.4个
题型:单选题难度:中档来源:不详

答案

由于 P(x,y)是椭圆 
x2
a2
+
y2
b2
=1
(a>0,b>0)上的点,则
x2
a2
+
y2
b2
=1

①(a2+b2)=(a2+b2)(
x2
a2
+
y2
b2
)
≥(x+y)2,故①正确;
(class="stub"1
x2
+class="stub"1
y2
)(
x2
a2
+
y2
b2
)≥(class="stub"1
a
+class="stub"1
b
)
2
,故②也正确;
③由椭圆的参数方程知
a2
x2
+
b2
y2
=class="stub"1
sin2x
+class="stub"1
cos2x
=class="stub"1
sin2x•cos2x
=class="stub"4
sin22x
,显然③也正确;
④由于Q(x′,y′) 是椭圆 
x2
a2
+
y2
b2
=1
(a>0,b>0)上的点.
依据椭圆的有界性知xx′≤a2,yy′≤b2,故class="stub"xx′
a2
+class="stub"yy′
b2
≤1
,故④也正确.
故答案选D.

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