直线l：y=1/2x+m与椭圆^交于不同两点A,B,圆O上存在两点C,D,CA=CB DA=DB求CD/AB取最小值时直线l的椭圆方程：x^2/16+y^2/4=1 圆O：x^2+y^2=4 CA=CB DA=DB

Question

直线l：y=1/2x+m与椭圆^交于不同两点A,B,圆O上存在两点C,D,CA=CB DA=DB求CD/AB取最小值时直线l的
椭圆方程：x^2/16+y^2/4=1 圆O：x^2+y^2=4 CA=CB DA=DB

Answer 1

直线方程代入椭圆方程求交点,x^2/16+(1/2x+m )^2/4=1,x^2+2mx+2m^2-8=0,
x1=-m+√(8-m^2),x2=-m-√(8-m^2),y1=1/2*( m+√(8-m^2)),y2=1/2*( m-√(8-m^2))
设A(x1,y1),B(x2,y2),则其垂直平分线为：y=(x1-x2)/(y2-y1)*(x-1/2*(x1+x2))+1/2*(y1+y2)=4(x+m)+m=4x+5m,与圆相交,x^2+(4x+5m )^2=4,17x^2+40mx+25m^2-4=0,
x3=(-20m+2√(68-25m^2))/17,x4=(-20m-2√(68-25m^2))/17,
y3=(5m+8√(68-25m^2))/17,y4=(5m-8√(68-25m^2))/17
则C(x3,y3),D(x4,y4)
|AB|=√((x1-x2)^2+(y1-y2)^2) =√(2√(8-m^2))^2+(√(8-m^2))^2) =√5*(8-m^2),
|CD|=√((x3-x4)^2+(y3-y4)^2)= √((4√(68-25m^2))/17)^2+(16√(68-25m^2))/17)^2)
= 4√(68-25m^2))/√17,
令|CD|/|AB|= 4√(68-25m^2)) /[√5*(8-m^2)*√17]=s,
s对m求导,s’=-4/√85*m(25m^2+64)/[(8-m^2)^2*√(68-25m^2)],
令s’=0,则m=0,在此时有极小值,因此,直线l的方程为y=1/2*x