Question

已知椭圆C:x2/a2+y2/b2=1(a>b>0)的离心率1/2,以原点为圆心,椭圆的短半轴为半径的圆与直线x-y+根

Answer 1

设A(m,n),B(m,-n),BP:x=ky+4,代入椭圆方程,得(3k^2+4)y^2+24ky+36=0

y1=-n,y2=y1y2/y1=36/[-n(3k^2+4)],结合k=(4-m)/n,得y2=36/[-n( (3(m-4)^2+4n^2)/n^2]

又m^2/4 + n^2/3=1,得3m^2+4n^2=12,故y2=3n/(2m-5)

E(x2,y2),x2=ky2+4=(4-m)/n*3n/(2m-5)+4=(5m-8)/(2m-5)

AE:(y-n)/(x-m)=(y2-n)/(x2-m)=n(m-4)/[(m-1)(m-4)]=n/(m-1)

令Y＝0,得x=1,即恒过Q(1,0)请采纳x05x05谢谢!