已知椭圆x²/a²+y²/b²=1(a>b>0)经过点M(√6,1)离心率为√2

1个回答

  • (1)椭圆过点(√6,1),

    ∴6/a^2+1/b^2=1.

    离心率c/a=1/√2,∴a^2=2c^2,b^2=c^2,代入上式得4/c^2=1,c^2=4,a^2=8,

    ∴椭圆的标准方程是x^2/8+y^2/4=1.

    (2)设A(2√2cosu,2sinu),B(2√2cosv,2sinv),则

    向量PA*PB=(2√2cosu-√6,2sinu-1)*(2√2cosv-√6,2sinv-1)

    =(2√2cosu-√6)(2√2cosv-√6)+(2sinu-1)(2sinv-1)

    =8cosucosv-4√3(cosu+cosv)+6+4sinusinv-2(sinu+sinv)+1=-2,

    ∴8cosucosv+4sinusinv-4√3(cosu+cosv)-2(sinu+sinv)+9=0,①

    AB的斜率=(sinu-sinv)/[√2(cosu-cosv)],

    AB的方程是y-2sinu=(sinu-sinv)(x-2√2cosu)/[√2(cosu-cosv)],

    设AB过定点(m,n),则√2(cosu-cosv)(n-2sinu)=(sinu-sinv)(m-2√2cosu),

    ∴√2(cosu-cosv)n-(sinu-sinv)m+2√2(sinucosv-cosusinv)=0.②

    不存在常数m,n,使得方程①、②同解,

    ∴AB不过定点.