Question

已知椭圆x2/a2+y2/b2=1与直线L：X+Y-1=0相交于两点PQ且OP垂直于OQ,O为坐标原点 试证明：

Answer 1

P(x1,y1),Q(x2,y2),有：x1+y1-1=0,x2+y2-1=0 ...(1)

x+y-1=0代入x^2/a^2+y^2/b^2=1,得：

==> (a^2+b^2)x^2-2a^2*x+a^2(1-b^2）=0

x1+x1 =2a^2/(a^2+b^2),x1*x2=a^2(1-b^2)/(a^2+b^2) ...(2)

OP垂直OQ：(y1/x1)(y2/x2)=-1 ...(3)

(1)(2)(3) ==> 1/a^2 +1/b^2 = 2 = 定值 ...(4)

e=c/a=[√(a^2-b^2)]/a =√[1-(b/a)^2]

(√2)/2≥e=√[1-(b/a)^2]≥(√3)/3 ...(5)

(4)(5) ==> (√6)/2≥a≥(√5)/2