Question

椭圆x2+4y2=16上有两点P、Q,O为原点,若OP、OQ斜率之积为 -1/4,求证|OP|2+|OQ|2为定值20.

谢谢各位啦

Answer 1

设两点(x1,y1),(x2,y2),斜率分别是k1,k2
则k1k2=y1y2/x1x2=-1/4
根据
X^2/16+Y^2/4=1
y^2=4-x^2/4
所以[sqrt(4-x1^2/4)*sqrt(4-x2^2/4)]/x1x2=-1/4
可以化得
x2^2=16-x1^2
|OP|^2+|OQ|^2=x1^2+y1^2+x2^2+y2^2=4-(3/4)*x1^2+4-(3/4)*x2^2
=20

定值20,与P,Q的坐标无关
ps:sqrt是根号的意思

Answer 2

证明:
设P(4cosα,2sinα),Q(4cosβ,2sinβ).
∵kOP×kOQ=-1/4,
∴(2sinα/4cosα)×(2sinβ/4cosβ)=-1/4
→cos(α-β)=0,
即α-β=2kπ±π/2,k∈Z.
所以,
|OP|^2+|OQ|^2
=16(cosα)^2+4(sinα)^2+16(cosβ)^2+4(sinβ)^2
=20(cosβ)^2+20(sinβ)^2
=20为定值

Answer 3

难啊