椭圆x的平方除以a的平方+y的平方除以b的平方=1(a>b>0)与直线x+y-1=0交于M,N两点 且以MN为直径的圆过
1个回答

(1)

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

b^2x^2+a^2y^2=a^2b^2

x+y-1=0

y=-x+1

将y=-x+1,代入b^2x^2+a^2y^2=a^2b^2,

b^2x^2+a^2(-x+1)^2=a^2b^2,

即(b^2+a^2)x^2-2a^2x+a^2-a^2b^2=0,

x1+x2=2a^2/(b^2+a^2),

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

∵OP⊥OQ,

∴x1*x2+y1*y2=0,

即x1*x2+(x1-1)*(x2-1)=0

∴2x1*x2-(x1+x2)+1=0,

∴2a^2-2a^2b^2-2a^2+b^2+a^2=0,

得,

b^2+a^2=2a^2b^2,

即1/a^2+1/b^2=2,

(2)

e^2=1-a^2/b^2=1+1/(1-2a^2),

∵1/2≤e^2≤3/4,

∴3≤2a^2≤5,

∴长轴范围是√6≤2a ≤√10