Question

已知双曲线C:x2/a2-y2/b2=1(a>0,b>0)离心率为√3,顶点到渐近线的距离为√6/3 一求双曲

Answer 1

顶点是(a,0),渐进线方程是bx+ay=0

故有|ab|/根号(b^2+a^2)=ab/c=根号6/3

又e=c/a=根号3

故b=根号6/3*根号3=根号2

c^2=a^2+b^2

3a^2=a^2+2

a^2=1

故方程是x^2-y^2/2=1.

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,上式分子为0,

XaXb+YaYb ＝ 0

向量OA和OB垂直,∠AOB = 90度