Question

椭圆X^2/a^2+y^2/b^2=1（a＞b＞0）与直线x+y=1交与P,Q两点且OP垂直于OQ,其中O为坐标原点 求

Answer 1

x+y=1,x=1-y,y=1-x

椭圆X^2/a^2+y^2/b^2=1

OP垂直于OQ,交点 P(0,1),Q(1,0)

X^2/a^2+(1-x)^2/b^2=1;x=0 ==>1/b^2=1

(1-y)^2/a^2+y^2/b^2=1;y=1 ==> 1/a^2=1

1/a^2+1/b ^2=2