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