已知直線y=ax+1與雙曲線3x^2-y^2=1交于A,B兩點
(1)求a的取值范圍
(2)若以AB為直徑的圓過坐標(biāo)原點,求實數(shù)a的值
(3)若S△OAB=2,求a的值
(4)是否存在這樣的實數(shù)a,使A,B兩點關(guān)于直線y=x/2對稱?若存在,請求出a的值,若不存在,說明理由.
將y=ax+1代入3x^2-y^2=1,得：(3-a^2)x^2 -2ax -2 =0
1.交于A,B兩點 ==> 判別式 =(2a)^2 -4*(3-a^2)(-2)> 0
==> -根號6 < a < 根號6
2.以AB為直徑的圓過坐標(biāo)原點
OA垂直O(jiān)B,(y1/x1)(y2/x2) = -1,x1x2+y1y2=0
x1x2+(a*x1 +1)(a*x2 +1) =0
(1+a^2)x1x2 +a(x1+x2)+1 =0 ...(1)
x1x2 =-2/(3-a^2),x1+x2 =2a/(3-a^2)代入(1),得:
a = 1,-1
3.2 = S△OAB =AB*d/2,d =1/根號(1+a^2) = 點O到直線AB距離
AB = |x1-x2|*根號(1+a^2)
==> |x1-x2| =4,(x1+x2)^2 -4x1x2 =16 ...(2)
x1x2 =-2/(3-a^2),x1+x2 =2a/(3-a^2)代入(2),解得:
a = 根號2,-根號2,(根號15)/2,-(根號15)/2
4.A,B兩點關(guān)于直線y=x/2對稱
A,B的中點在直線y=x/2上:
(y1+y2)/2 =[(x1+x1)/2]/2,a(x1+x2)+2 =(x1+x2)/2
x1+x2 =2a/(3-a^2)代入,解得:a=6 > -根號6
因此,不存在實數(shù)a,A,B兩點關(guān)于直線y=x/2對稱