椭圆X^2/25+y^2/5=1上有两点P,Q.O为坐标原点,且直线OP,OQ斜率之积为1/5,求证OP^2+OQ^2为定值

设P(x1,y1) Q(x2,y2)。直线的斜率可以求导数得到，具体过程如下：
白敲了，悲剧

椭圆方程为：(x/5)^2+(y/√5)^2=1
设P(5cosα,√5sinα)，Q(5cosβ,√5sinβ)
则Kop=(√5/5)tanα，Koq=(√5/5)tanβ
所以，Kop*Koq=(1/5)tanαtanβ=1/5
===> tanαtanβ=1
===> α+β=π/2
所以，cosα=sinαβ
OP^2+OQ^2=25cos^2 α+5sin^2 α+25cos^2 β+5sin^2 β
=5+20cos^2 α+5+20cos^2 β
=10+20(cos^2 α+cos^2 β)
=10+20(sin^2 β+cos^2 β)
=10+20
=30

设两点P(x1,y1),Q(x2,y2),斜率分别是k1,k2
则k1k2=y1y2/x1x2=1/5
根据
X^2/25+Y^2/5=1
y^2=5-x^2/5
所以[根号(5-x1^2/5)*根号(5-x2^2/5)]/x1x2=1/5
根号(25-x2^2-x1^2+x1^2x2^2/25)=x1x2/5
可以化得
x2^2=25-x1^2
|OP|^2+|OQ|^2=x1^2+y1^2+x2^2+y2^2=5+(4/5)*x1^2+5+(4/5)*x2^2
=10+4/5*25
=10+20
=30