Question

希望十二小时内有解答已知双曲线c:x^2/a^2-y^2/b^2=1(a>0,b>0)的离心率为√3,右准线方程x=√3

Answer 1

这种题 你就不能怕麻烦,就得死算.

（1） e=c/a = √3,a^2/c =√3/3

a=1,c = √3,b =√2,双曲线方程为

2x^2 -y^2 = 2

x^2+y^2=2上动点P(x0,y0)(x0y0≠0)处的切线方程为

x0x+y0y = 2

(2)

设A,B,的坐标为（Xa,Ya),(Xb,Yb),

则（Xa,Ya),(Xb,Yb) 为方程组

x0x+y0y = 2 (1)

2x^2-y^2 = 2 (2)

的解

(1) 代入（2）消去y,得到

(2-x0^2/y0^2)/x^2 +4x0x/y0^2 - (4/y0^2+2) = 0

XaXb = - (4/y0^2+2)/(2-x0^2/y0^2) = -(4+2y0^2)/(2y0^2-x0^2)

(1) 代入（2）消去x,得到

(2y0^2-x0^2)y^2 -8y0y + 8-2x0^2 = 0

YaYb = (8-2x0^2)/(2y0^2-x0^2)

XaXb+YaYb

= -(4+2y0^2)/(2y0^2-x0^2) + (8-2x0^2)/(2y0^2-x0^2)

= [4-2(x0^2+y0^2)]/(2y0^2-x0^2)

(x0,y0) 是圆x^2+y^2=2的点,上式分母为0,

XaXb+YaYb ＝ 0

向量OA和OB垂直,∠AOB = 90度