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直线Y=mX+1(m大于0)与椭圆2X^2+Y^2=2相交于A,?

2007-02-07 13:13:301***
直线Y=mX+1(m大于0)与椭圆2X^2+Y^2=2相交于A,B两点。:请参阅附件.?

最佳回答

  • 请参阅附件.
    2007-02-09 16:51:05
  • 1. 把Y=mX+1代入2X^2+Y^2=2得,(2+m^)x^+2mx-1=0,A(x1,y1),B(x2,y2),则x1+x2=-2m/(2+m^),x1x2=-1/(2+m^),|AB|=√(1+m^)[(x1+x2)^-4x1x2]=(6√2/5)^,解得m^=1/2,∴m=√2/2 2. 点(0,0)到AB的距离d=1/√(1+m^),|AB|=√(1+m^)[(x1+x2)^-4x1x2],|AB|×d=4/3,∴[(x1+x2)^-4x1x2]4/3,∴m^=1+√3,m=√(1+√3),设P(x,y),AB,OP交点为M(x',y'),则x'=(x1+x2)/2=-m/(2+m^),y'=mx'+1=2/(2+m^),∴x=2x'==-2m/(2+m^)=-2√(1+√3)/(3+√3),y=2y'=4/(2+m^)=4/(3+√3)
    2007-02-09 16:47:38
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