Question

【数学填空题求解】m的取值范围是

已知圆G:X^2+Y^2-[2根号2]x-2y=0经过椭圆x^2/a^2+y^2/b^2=1[a>b>0]的右焦点及上顶点，过椭圆外一点M(m,0),(m>a),倾斜角为120°的直线l交椭圆于C,D两点，若点N(3，0)在以线段CD为直径的圆E的外部，则m的取值范围是

Answer 1

圆G:X^2+Y^2-2√2x-2y=0经过椭圆x^2/a^2+y^2/b^2=1(a>b>0)的右焦点(c,0)及上顶点(0,b),
∴c^2-2√2c=0,b^2-2b=0,b>0,c>0,

∴c=2√2，b=2,
∴a^2=b^2+c^2=12,
∴椭圆方程是x^2/12+y^2/4=1,①
把l:y=-√3(x-m),②
代入①，x^2+9(x^2-2mx+m^2)=12,
整理得10x^2-18mx+9m^2-12=0,
△/4=81m^2-10(9m^2-12)=120-9m^2>0,m^2<120/9,-2√30/3<m<2√30/3，③
设C(x1,y1),D(x2,y2),则x1+x2=9m/5,x1x2=(9m^2-12)/10,
由②，y1+y2=-√3(x1+x2-2m)=√3m/5,
y1y2=3(x1-m)(x2-m)=3[x1x2-m(x1+x2)+m^2]
=3[9m^2-12-18m^2+10m^2]/10=3(m^2-12)/10,
圆E:(x-x1)(x-x2)+(y-y1)(y-y2)=0,
即x^2+y^2-9mx/5-√3my/5+(12m^2-48)/10=0,
点N(3，0)在圆E外部，
∴9-27m/5+(6m^2-24)/5>0,

整理得2m^2-9m+7>0,
解得m<1或m>7/2,
由③，-2√30/3<m<1或7/2<m<2√30/3，为所求.