Question

已知曲线C1的方程为x^2-y^2/8=1(x>=o,y>=0),圆C2的方程为（x-3）^2+y^2=1,斜率为k(k

Answer 1

C2的圆心为C(3,0),半径为r = 1

x² - y²/8 = 1,x ≥ 0,y ≥ 0为双曲线在第一象限的部分.

|AC| = r = 1

|AB| = √3

AC⊥AB,|BC|² = |AC|² + |AB|² = 1 + 3 = 4

即B为圆(x - 3)² + y² = 4和C1的交点.

消去y得:(3x + 1)(x - 1) = 0

x = 1 (舍去x = -1/3)

B(1,0)

切线:y - 0 = k(x - 1),kx - y - k = 0

C与切线的距离d = r = 1 = |3k - 0 - k|/√(k² + 1)

3k² = 1

k = √3/3 (舍去k = -√3/3 < 0)