已知圆O:x^2+y^2=1,O为坐标原点,一条直线l:y=kx+b(b>0)与圆O相切并与椭圆x^2/2+y^2=1交

1个回答

  • (1)kx-y+b=0

    b/√(1+k2)=1

    f(k)=√(1+k2)

    (2) A(x1,y1) B(x2,y2)

    x1x2+y1y2=2/3

    把y=kx+b代入x^2/2+y^2=1

    (1+2k2)x2+4bkx+2b2-2=0

    x1x2=(2b2-2)/(1+2k2)=2k2/(1+2k2)

    x1+x2=-4bk/(1+2k2)

    y1y2=k^2x1x2+kb(x1+x2)+b2

    =2k^4/(1+2k2)-4k2b2/(1+2k2)+b2

    所以

    2k^4/(1+2k2)-4k2b2/(1+2k2)+b2+2k2/(1+2k2)=2/3

    解得k=±1 ,b=√2

    所以直线l的方程y=±x+√2

    (3)

    2k^4/(1+2k2)-4k2b2/(1+2k2)+b2+2k2/(1+2k2)=m

    (k2+1)/(2k2+1)=m

    2/3≤m≤3/4

    2/3≤(k2+1)/(2k2+1)≤3/4

    得1/2≤k2≤1

    又由(1+2k2)x2+4bkx+2b2-2=0

    x1x2=(2b2-2)/(1+2k2)=2k2/(1+2k2)

    x1+x2=-4bk/(1+2k2)

    得|AB|=2√2√(1+k2)√k2/(1+2k2)

    高为1

    所以面积S=1/2*2√2√(1+k2)√k2/(1+2k2)

    因为1/2≤k2≤1

    所以S的范围为