Question

已知双曲线X^2-Y^2=1,椭圆与该双曲线共焦点,且经过点（2,3）,求（1)椭圆的左右顶点分别是A,B,右焦点

Answer 1

（1）双曲线X^2-Y^2/3=1的焦点是（土2,0),

设所求椭圆方程是x^2/(b^2+4)+y^2/b^2=1,

它过点(2,3),

∴4/(b^2+4)+9/b^2=1,

4b^2+9b^2+36=b^4+4b^2,

b^4-9b^2-36=0,b^2>0,

∴b^2=12,

∴椭圆方程是x^2/16+y^2/12=1.其左右顶点分别是A(-4,0),B(4,0),右焦点F(2,0),N(8,n),

AN:y=n(x+4)/12与椭圆交于M(4(108-n^2)/(108+n^2),72n/(108+n^2)),

AM=MN,

[4(108-n^2)/(108+n^2)+4]^2+[72n/(108+n^2)]^2

=[4(108-n^2)/(108+n^2)-8]^2+[72n/(108+n^2)-n]^2,

∴12[8(108-n^2)/(108+n^2)-4]+n[144n/(108+n^2)-n]=0,

48(108-3n^2)+n^2(36-n^2)=0,

n^4+108n^2-48*108=0,

∴n^2=36,n=土6.M(2,土3）,

当M(2,3)时MA=(-6,-3),MB=(2,-3),|MA|=3√5,|MB|=√13,MA*MB=-3,

cosAMB=-√65/65,

当M(2,-3)时cosAMB=-√65/65.

（2）?